For my medical reblogs

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Jul 5
Tidal percussion

Tidal percussion

Microbiology Genetics

itisoktobesmart:

  1. Transformation
  2. Conjugation:  F+ —> F-  &  Hfr —> F-
  3. Transposition 
  4. Transduction: Generalized   &  Specialized

(Source: sidratimes)

Slipped disc

Slipped disc

BASIC MOTOR PATHWAY

Corticospinal tracts

sub arachnoid hemorrhage

sub arachnoid hemorrhage

medicowesome:

Minimum alveolar concentration, blood gas partition coefficient, AV gradient of anaesthetics simplified

Hi!
This is what I understood. I hope it helps you understand as well =)

I talk about minimum alveolar concentration, blood gas partition coefficient, AV gradient of anaesthetics, rise in partial pressure, lipid and blood solubility, potency, time required for induction and amount of gas required to saturate blood.

I am just a medical student and not a pro anesthesiologist or anything. If I have made a mistake or explained a concept wrong, lemme know!

Anteroinferior dislocation of shoulder injuries axillary nerve 

Cholestyramine

nakimedicalblog:

The most common side effect of cholestyramine is?

1) Flatulent diarrhoea
2) Constipation
3) Tooth decay
4) Itching
5) Black stools

It causes constipation. Also used in post-ileal resection Crohn’s patients.

The Crohn’s disease concept is so cool.

Cholestyramine is a bile acid sequestrant that binds bile salts, which are poorly absorbed in Crohn’s disease and might otherwise cause colonic irritations and diarrhea upon entering the colon.

Cholestyramine

The most common side effect of cholestyramine is?

1) Flatulent diarrhoea
2) Constipation
3) Tooth decay
4) Itching
5) Black stools

Excitotoxicity, defined as excessive exposure to the neurotransmitter glutamate or overstimulation of its membrane receptors, has been implicated as one of the key factors contributing to neuronal injury and death in a wide range of both acute and chronic neurologic disorders. Excitotoxic cell death is due, at least in part, to excessive activation of N-methyl-d-aspartate (NMDA)-type glutamate receptors and hence excessive Ca2+ influx through the receptor’s associated ion channel. 

Excitotoxicity, defined as excessive exposure to the neurotransmitter glutamate or overstimulation of its membrane receptors, has been implicated as one of the key factors contributing to neuronal injury and death in a wide range of both acute and chronic neurologic disorders. Excitotoxic cell death is due, at least in part, to excessive activation of N-methyl-d-aspartate (NMDA)-type glutamate receptors and hence excessive Ca2+ influx through the receptor’s associated ion channel. 

Stringency is the extent to which hybridisation can occur between nucleic acids with mismatched sequences. High-stringency conditions require absolute complementarity between the molecules, while low-stringency conditions permit hybridisation where there are some mismatched bases. Nucleic acids that are perfectly matched, under high-stringency conditions, are said to be homologous; nucleic acids that hybridise despite some base mismatches, under low-stringency conditions, are said to beheterologous.

Typically, high-stringency conditions are achieved either by reducing NaCl concentration or increasing temperature, close to the melt temperature (Tm) of the molecules involved. Lower-stringency conditions are achieved by increasing NaCl concentration or decreasing temperature considerably below the Tm of the molecules. The level of stringency can be varied by adjusting temperature, salt concentration or formamide concentration; this is important when, for example, searching for a single-base mutation in a DNA sequence. 

Genetics

If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous(Ss) for the sickle-cell gene

p is 1 , but always we shud think of the small numbers to get ti rite:i mean always consider : p=1-q (q is the ss here , and is generally very small….but impo to get it rite):
q2=.09 = ss 
q = Square root of .09 = .3
p = 1 - .3 = .7
2pq = 2 (.7 x .3) = .42 = 42% of the population are heterozyotes (carriers)

2. A patient presents to the physician’s office to ask questions about
color blindness. The patient is color-blind, as is one of his brothers. His
maternal grandfather was color-blind, but his mother, father, daughter, and
another brother are not. His daughter is now pregnant. The risk that her
child will be color-blind is
a. 100%
b. 50%
c. 25%
d. 12.5%
e. Virtually 0
In this Question, all men are affected and no male transmitted the disease to another male 

so X-linked recessive it is

Males need just one x to get affected
Females need xx to be affected

2. So the daughter is xX ( heterozygous carrier) 

She marries a normal male (XY). Now the second step is to do Punett square

xX vs XY 

which gives us XX XY Xx & xY = normal femae, normal male, carrier female & affected male 

so 25% is the answer, simple !

3. blue eyes is a recessive trait. It is said that in a given population 1/100 have blue eyes. How many people are carrier for the gene?
1/100
9/100
10/100
18/100…..answer
4. a man who is a known heterozygous carrier of albinism marries his half first cousin i.e they share a common grandparent….the trait is transmitted as autosomal recessive…wat is the probability tat this couple will produce child with this disorder..

1.1/2
2.1/4
3.1/8
4.1/16
5.1/32
6.1/64
answer is 1/64..its in kaplan genetics book…they say tat probalility goes as follows…
first for the herterozygous carrier man
grandparent to fther=1/2 n then parent to the man in picture above=1/2..tat cunts to 1/4
same way grandparent to father of the grl=1/2..n then father to tat grl=1/2…tats 1/4

nw both of then marry so probablity nw is 1/4*1/4=1/16
n nw tt they will produde son with this disorder =1/16*1/4==1/64

i got the answer bt nw i imagine…in kaplan it says tat first cousins share 1/8th of their genrs….so shudnt it b 1/8*1/8*1/4=1/256 the answer

is it nt…from grandparent to parent =1/2..nw transmission of same affected gene frm parent to son=1/4..i. e half the probablity of father…if u do in tat manner then its k tat first cousins share 1/8th of their genes

5. albinism is an autosomal recesiive disease. The frequency of the gene for albinism in a population is know to be a 1/190. What is the aproximate risk for albinismin the child of an albino woman who marries an unrelated and unaffected man whit no family history of albinism?
a. 1/95
b. 1/190
c. 1/380
d. 1/570
e. 1/170
answer is b
gene frequency = 1/190
so carrier frequency = 2 x 1/190 
so chance of father being carrier is 2/190
mother is affected.. so chance of baby with albinism, for affected mother and carrier father is 1/2
so combined chance = 1/2 x 2/190 = 1/190

6. SC is an autosomal recessive conditin that occurs with a freq of 1/625 among African-americans. What is the approximate chance that the clinically normal sister of a person with sicle willhave an affected child if she marries and unrelated african american man?
a. 1/4
b. 1/25
c. 1/50
d. 1/75
e. 1/625
Answer is 1/75
An unaffecte woman whose brother has sicle has a 2/3 chance of being a carrier of the abnormal gene.
The diseas has a frequency of 1/625 =qsquare.
So the frequency of the abnormal gene (q) is 1/25
————————————————————————
According to the H-w law, the frequency of a carriers in the general population is 2pq means 2 x 1 x 1/25 = 2/25, wich is the chance that the woman’s husband carries the sickle gene.
The chance that the woman would have an affected child is determined by multiplying the chance that she is a carrier by the chance that her husband is a carried, multiplied by the chance that the child would be affected if both parents are carriers
2/3 x 2/25 /1/4 =4/300 =1/75

7. Cf is the most common autosomal recesive disease among white Amnericans, occuring at frecuency of 1/2500 births. What is the approximate frecuency of heterozygous carriers for cystic fibrosis in this population?

a. 1/25
b. 2/25
c. 1/50
d. 1/2500
e. 2/2500
Answer i/25
Hardy-Weinberg equation p^2 (AA), 2pq(Aa), q^2(aa) 
If the prevalence of CF is 1/2500, the frequency of a gene a (q) is square root of 1/2500=1/50
if carrier rate (Aa) is 2pq, then it’s 2 x 49/50 (p=50/50-1/50=49/50) x 1/50 (q) = 0.04=4% of the population, or a 1/25 carrier rate
8. Bunch of questions:
1. wich one of the following types of genetic diseas occurs more frequently?
AD
AR
X linked
chromosome abnormality
multifactorial

2. The overall incidence among livebirht of genetic disease and other congenital anomalies apparent by 25 years of age is
0.5%
1%
8%
15%

3. The proportion of adult hospital admision that occurs for idseases that are due mostly or entirely to genetic factors is approximately
1%
10%
15%
20%
25%
answers:1. multifactorial…wich an incidence about 46/1000.
2. 8%..although most individual genetic diseases are rare, genetic diseases and congenital anomalies as a gropu are quite common, affecting some 79/1000 individuals by age 25 years
3. 10% of hospital admissions are for diseases that are caused largely or entirely by genetic factors.

1. if a parent has an isochroosome for an aoutosome such as i(21q) what is the risk for trisomic offspring
10%
25%
50%
100%

1. 100%…if a parent has after the second miotic division 2 chromosomas of the same when it gets combined with the one of the other parent, there is 100% posibility of trisomy


2. The most common translocation found in humans with an incidence rate of 1/1500 is
45 xx 13 q 14 q
45 xy 14 q 21 q
2. “The mos common is 45 xx or 45 xy affecting 13 and 14 chromosomes.
The translocation among 14 and 21 is the second most common
The recurrence risk to robertsonian translocation carries of all types of abnormal offspring depends on ascertainment, the specific chromosomes involved, and the sex of the carrier parent.
Females whit translocation 14 - 21 have a risk for offspring with down syndrome of 10%.
Whereas the male carrier has a 2% risk

A 10-year-old boy with the facies and many other critical features of Down’s syndrome has an intelligence quotient (IQ) in the mid-normal range. Which of the following genetic mechanisms would most likely account for the discrepancy between the child’s IQ and his appearance?
A. Balanced translocation
B. Chiasma
C. Mosaicism
D. Spermiogenesis
E. Synapsis
The correct answer is C. Mosaicism is the term used when cells with more than one type of genetic constitution are present in the same organism. The scenario described in the question uncommonly occurs when nondisjunction of chromosome 21 occurs during mitosis (rather than meiosis) in one of the early cell divisions. The degree to which the individual expresses the characteristics of the syndrome depends on the number of cells involved and their distribution.

Balanced translocation (choice A) does not produce features of any syndrome, because critical genetic material is not lost, although progeny may be affected when the translocated chromosome is added to a complement of otherwise normal chromosomes.

Chiasma (choice B) refers to the “X”-shape of chromosomes undergoing exchange of genetic material in crossover.

Spermiogenesis (choice D) refers to the development of sperm precursors into mature sperm.

Synapsis (choice E) refers to the pairing of homologous chromosomes in the first meiotic division. 

A man is affected with an X-linked dominant disorder. The penetrance of the disease genotype is 60%. Assuming that he mates with a genetically unaffected female, what is the probability that his daughters will be affected with this disorder?
A. 0%
B. 30%
C. 50%
D. 60%
E. 70%
Answer is 60%

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* Re:To Elba 
#1267326
elbamaritza - 04/11/08 22:15

A 25-year-old woman has mild expression of hemophilia A. A genetic diagnosis reveals that she is a heterozygous carrier of a mutation in the X-linked factor VIII gene. Which of the following is the most likely explanation for mild expression of disease in this individual?
A. Her father is affected, and her mother is a heterozygous carrier
B. A high proportion of the X chromosomes carrying the mutation are active
C. A nonsense mutation is causing formation of a truncated protein
D. One of her X chromosomes carries the SRY gene
E. X inactivation does not affect the entire chromosome
B it is..

The correct answer is B. The most likely explanation for mild expression in a heterozygous carrier is that, when X inactivation occurred in the affected individual, the random process happened to inactivate most of the X chromosomes that carried the normal version of the factor VIII gene. Thus, most of the active X chromosomes in this individual would carry the mutation and would not produce functional factor VIII, leading to a clinically expressed deficiency.

If the woman’s father were affected and her mother were a carrier (choice A), she would have a 50% chance of being an affected homozygote, but her expression would more likely be severe.

A nonsense mutation (choice C) would likely produce severe expression, if it were inherited from both the mother and the father.

The SRY gene (choice D) is involved in sex determination and would not affect factor VIII expression.

Although it is true that X inactivation does not affect the entire X chromosome (choice E), it 
consistently affects the factor VIII gene and therefore this could not explain the status of this woman

Below is a short sequence of DNA showing the mutation that causes B-thalassemia. Which of the following DNA mutations causes B-thalassemia?

Normal B globulin allele ————- AAC CAG AGG
B glbulin allele (thalassemia) —- AAC TAG AGG

a) Frameshift
b) Insertion
c) Missense
d) Nonsence
e) Silent
5’-AAC TAG AGG-3’ this is the DNA sequence showing the strand which is not transcribed.
Consider the following:
5’-AAC TAG AGG-3’
3’-TTG-ATC-TCC-5’
The bottom strand is transcribed as follows:
3’-TTG-ATC-TCC-5’ DNA
5’-AAC-UAG-UGG-3’ RNA
Now you have UAG, a stop codon in the RNA. No translation beyond that, nonsense

Question 3. Which karyotype represents an individual with a balanced chromosome complement? 


A) 69, XYY 
B) 45, XX, der(14;21)(q10;q10) 
C) 46, XX, del(4p) 
D) 46, XY, +der(1)t(1;3)(p31;p21)-3 
E) 46, XY, dup(12)(q11.2;q21.2)
B) 45, XX, der(14;21)(q10;q10) 
Explanation: This is the karyotype of an individual with a balanced Robertsonian translocation between the long arms of chromosomes 14 and 21. This individual lacks one normal 14 and one normal 21 and has instead a long chromosome composed of the q arms of both chromosomes (note that there are only 45 centromeres - translocation of two acrocentric chromosomes results in the loss of one centromere). As the short arms of acrocentric chromosomes are very small and contain no essential genetic material, their loss does not result in a clinical phenotype. However, the offspring of these balanced individuals are at risk of inheriting missing or extra parts of the involved chromosomes

Question 9. Ned and Stacey are the parents of Mark, a child affected with a fully penetrant, autosomal recessive disorder that is easily diagnosed at birth, and occurs in the population with an incidence of 1/3600. Neither Ned nor Stacey has this disorder themselves. Their next child, Tony, is born without any apparent signs of the disease. Tony grows up and marries Maria, a woman with no known family history of the disorder. The chance that Tony and Maria’s first child will be affected with the same disorder that affects Mark is closest to which of the following numbers? 


A) 1/45

B) 1/120

C) 1/180 

D) 1/240

E) 1/360
C) 1/180 
Explanation: Three things must occur for Tony and Maria to have an affected child.
1. Tony must be a carrier for the disorder. As Tony doesn’t show any symptoms of the disorder (which is fully penetrant), he does not carry both copies of the mutation. Subsequently, there are three equally likely remaining genotype options for Tony - homozygous normal, heterozygous with the mutation received from his mother, or heterozygous with the mutation received from his father. The overall probability that Tony is a carrier is therefore 2/3.
2. Maria must also be a carrier. Because Maria has no family history of disorder, we assume her risk of being a carrier is equivalent to the carrier frequency in the population. This can be deduced using Hardy-Weinberg equilibrium. The incidence of this disorder in the population is 1/3600. As mentioned in the text (p. 126), a rough estimate of the carrier frequency can be obtained by doubling the square root of the population frequency - ≈ 2 × 1/60 = 1/30. This is the chance that Maria is a carrier.
3. Both Tony and Maria must pass their mutation on to a child. If both Tony and Maria are carriers, the chance they would have an affected child is 1/4.
To obtain the final probability for this problem, multiply the probabilities from all three events - 2/3 × 1/30 × 1/4 = 2/360 = 1/180.
Question 7. Type 1 albinism is an autosomal recessive disorder that results from a mutation in tyrosinase, resulting in the lack of pigmentation. John and Susan, who are married, are both healthy and not affected with type 1 albinism, although John’s father was affected as was Susan’s maternal grandmother. Susan is pregnant. What is the probability that this fetus will be affected with type 1 albinism? 


A) 1/4

B) 1/8 

C) 1/16

D) 1/24

E) 1/64
B) 1/8 
Explanation: Individuals with autosomal recessive disorders have mutations in each copy of the causative gene. For a child of John and Susan to have type 1 albinism, each parent must inherit one copy of the mutation and then pass it to the child. John’s father had type 1 albinism so John inherited one copy of the tyrosinase mutation from him. There is a 1/2 chance he will pass this along to a child. We know Susan’ mother is an obligate carrier for type 1 albinism (she had to inherit one of the two mutant copies from Susan’ grandmother, who was affected.) Subsequently, there is a 1/4 chance Susan will pass along a tyrosinase mutation to a child - a 1/2 chance she received the mutation her mother inherited from Susan’s grandmother multiplied by a 1/2 chance that she would pass along that mutation to a child. So there is a 1/8 total chance the fetus will inherit two copies of the tryrosinase mutation

Question 8. Suppose a research study shows that people who suffer from test anxiety are homozygous for a mutation in the EXAM gene. Individuals without test anxiety have the following sequence at the very beginning of the translated region of their EXAM genes:
5’-ATG ACG TTT GAA ATT CAG TCT AGA-3’
Met Thr Phe Glu Ile Gln Ser Arg
Affected individuals have the following sequence:
5’-ATG ACG TTT GAA ATT TAG TCT AGA-3’
Met Thr Phe Glu Ile STOP
The mutation identified among affected individuals is most likely an example of which of the following types? 


A) Missense 
B) Gain of function 
C) Nonsense 
D) Frameshift 
E) Dominant negative
C) Nonsense 
Explanation: A nonsense mutation is a substitution that leads to the generation of one of the stop codons, resulting in premature termination of the translation machinery

Question 10. Anne is 16 weeks’ pregnant and undergoes maternal serum screening to test her maternal serum alpha fetoprotein levels (MSAFP). Her MSAFP level is 4.5 multiples of the median (MoM). Assuming Anne is carrying a single fetus and her pregnancy is correctly dated for gestational age, for which of the following disorders should Anne be offered additional testing? 


A) Trisomy 13 
B) Pulmonary hypoplasia 
C) Open spina bifida 
D) Cystic fibrosis 
E) Trisomy 21
C) Open spina bifida 
Explanation: Increased levels of MSAFP are associated with the presence of open neural tube defects, such as anencephaly. The generally accepted cut-off level above which further investigation is offered is 2.5 MoM. This value detects approximately 75% of screened open neural tube defects .

Question 4. Robin is affected by an autosomal dominant disorder inherited from her mother. She is married to Chad, who is unaffected and has no history of the disorder in his family. Robin and Chad have two unaffected children. Studies suggest that for every 100 individuals who inherit mutations in the gene of interest, only 50 actually show clinical symptoms. The new mutation rate for this disorder is essentially zero. Based on this, what is the probability that their next child will present with the clinical signs of the disease. 


A) 3/4 
B) 1/2 
C) 1/4 
D) 1/8 
E) Virtually 0 
C) 1/4 
Explanation: We can assume that Robin carries one copy of the mutation. The probability that she passes the mutation to the next pregnancy is 1/2 and the probability that a fetus with the mutation will express clinical symptoms is also 1/2. Since these are independent events, they can be multiplied to obtain the answer of 1/4.

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* Re:To Elba 
#1267329
elbamaritza - 04/11/08 22:17


family with one child with downs is concerned of having another child with downs, so they approach to a specialist.DNA of both parents and the affected child are studied.A cloned DNA fragment when used as probe reveals RFLP in a region adjacent to centromere of ch21.Four haplotypes exist A,B,C,D.fATHER has AC genotype,mother has BD genotype ,while the affected child has ABC genotype.When did the nondisjunction occured??
A Meiosis I of father
B Meiosis II of father
C Meisis I of mother
D meiosis II of mother.

Answer is A ?????

A 14-year-old girl presents with a swollen left knee. Her parents state she suffers from haemophilia and has been treated for a right-sided haemarthrosis previously. What other condition is she most likely to have? 
A.A Turner’s syndrome 
B.A Down’s syndrome 
C.A Ataxia telangiectasia 
D.A Hunter’s syndrome
E.A Coeliac disease

Haemophilia is a X-linked recessive disorder and would hence be expected only to occur in males. As patients with Turner’s syndrome only have one X chromosome however, they may develop X-linked recessive conditions

1) A 60 year-old-male with painless hematuria is found to have a right-side renal mass. There is not significantly family history. Cytologic evaluation of the mass shows malignant cell with chromosome 3p deletion. The deletion most likely involves which of the following genes?

A) RB

B) VHL

C) NF-1

D) WT_1

E) BRCA-1
Patients with both sporadic and hereditary (associated VHL disease )renal cell carcinoma are found to have deletions of the VHL gene on chromosome 3p( short arm of chromosome 3).

Achondroplasia shows autosomal dominant inheritance with complete penetrance. A man with achondroplasia has an unaffected partner of normal height. Their first two children are not affected. Which of the following is correct?


a) The probability that their next child will be affected is 1 in 2.

b) The probability that each of their children will be affected is less than 1 in 2 because the parents of the man with achondroplasia are not affected.

c) The mother of these children has a sister with achondroplasia. This increases the risk that the children could be carriers.

d) If their third child is unaffected, then the probability that their fourth child will be affected is 1 in 8.


Answer A is correct

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* Re:To Elba 
#1267333
elbamaritza - 04/11/08 22:17

Tay-Sachs disease shows autosomal recessive inheritance. Parents of a newly diagnosed affected child are referred for genetic counseling. It would be correct to tell them that: 


a) the probability that their next child will be affected is 1 in 2.

b) the probability that the older unaffected sister of the affected child is a carrier is 1 in 2.

c) the fact that their last child was affected means that their next three children will not be affected.

d) the probability that each parent is a carrier is 1.
Aa x Aa= AA, Aa, Aa, aa,

So their next child will have 1/4 chance get affected.
older unaffected child can’t be an aa, right? So his/her genotype can onoy be AA, Aa, Aa, then the chance of his/her being a carrier is 2/3, not 1/2, since you can’t include the unexisting aa genotype in the denominator.
C is definetely wrong.

Neurofibromatosis type 1 is one of the most common autosomal dominant disorders. A woman with neurofibromatosis type 1 has an unaffected partner. Which of the following is correct regarding their children?


a) The probability that each of their children will be affected is 1 in 4.

b) The probability that their second child will be affected if their first child is affected is 1 in 4.

c) The probability that their third child will be affected if their first two children are affected is 1 in 2.

d) If their first child is affected then their second child will not be affected
Answer is cc

A molecular genetics group proposes to clone the region
of chromosome 17p11 to which the gene defective in
neurofibromatosis has been mapped. Both genomic
DNA and cDNA libraries are being considered for the
cloning procedure. An advantage of the genomic library
is that
(A) cloned DNA will include potential regulatory
elements
(B) 125I-labeled antibody to the protein product can be
used to probe the library
(C) introns will be absent in the cloned DNA
(D) only genes expressed in the cells will be cloned
(E) restriction endonucleases are not required in producing
the library
Genomic libraries originate by
cloning restriction fragments of DNA. They include all
elements associated with genes, such as promoters,
enhancers, and introns, that are involved in controlling
the expression of the gene

A couple seeks genetic counseling following the recent diagnosis of cystic fibrosis in their youngest child. Their two older children are healthy and have normal sweat chloride test results. Which of the following is the likelihood that each of the unaffected children is a carrier of cystic fibrosis? 

A) 1 in 2 
B) 1 in 3 
C) 1 in 4 
D) 2 in 3 
E) 3 in 4
for AR…..since there is one affected child father and mother both must be heterozygote…if u draw the punnet squire….than one will be affected,2 will be carrier and one will be unaffected…since we know one is affected..out of rest 3 2 will be carrier….so 2/3

About 5% of individuals in a particular population are known to carry a recessive gene for poliodystrophy. an inherited disorder 
characterized by recurrent seizures and dementia with onset in early childhood. A 32-year-old healthy woman who had a brother with this disorder seeks genetic counseling. The patient’s husband, an only child, does not know if his family has a history of the disorder. What is the probability that the patient is a carrier of poliodystrophy? 

A) 1/20 
B) 1/10 
C)3/8 
D)2/3 
E) ¾
this disease is AR…brother is affected….so both father and nother are carrier….chance of her being carrier would be 2/3….u exclude the brother as he is affeted…out of 3 left 2 will be acrrier and 1 will be unaffected….so 2/3

14 yo boy has haemophilia A,his older sister who does not have this condition ,ask abouth the risk that her own children will b affected

1/4
1/2
1/16
1/8
1/32


Answer is 1/8
disease is X linked recessive. Mother is most likely a carrier XX*, where X* is the diseased alele. So, 50% of her daughters will get the X*. that means chances that the girl is carrier is 1/2.
Now, chances the girl will give the X* to her son is 1/2 (and only sons will be affected). 
Now, chances she’ll have a son is 1/2.

So , overall P is 1/2*1/2*1/2=1/8

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* Re:To Elba 
#1267334
elbamaritza - 04/11/08 22:18

couple who have no children seek genetic counseling because the husband’s two brothers died in their teens of duchenne’s. Which of the following describes the husband’s risk of being a carrier of DMD? 
a) 0 
b) 1/4 
c) 1/3 
d) 1/2 
e) ¾

answer is A…dmd is an autosomic dominant with 100% penetrance, so if he is not sick, he does not have the gene.

a child knows to have Tay-sach’s disease, what is the probablity his Grandfather is a carrier? 

A. 0% 
B. 25% 
C. 50% 
D. 75% 
E. 100%

answer is c

its 50%..coz his father is a carrier… n their is 50% chance that he received the recessive gene from his father(grand father0..or his mother(grand mother)

A man who has Neurofibromatosis type 1 (autosomal dominant) marries a phenotypically normal woman. If they have five children, what is the probability that none of the children will be affected with this disorder? What is the probability that all of the children will be affected

to calculate this type of probability one child = 50%
father = Nn and the mother = nn
children = Nn Nn nn nn, then we have to calculate that all the children to be deseased:
1 child = 1/2, all five = (1/2)^ 5 or 1/2*1/2*1/2* 1/2* 1/2 = 1/32 =0.03125 or 3.125%.

Suppose that you have done a carrier test for PKU (autosomal recessive disorder) in a population, and you discover that the heterozygote carrier frequency is 1/500. Based on this information, what proportion of the population will be affected with PKU?

its application of hardy weinberg equillibrium….carrier iis 1/500..so 2pq is 1/500…here p is negligible…so q is 2q….q is 1/1000…q is genetic frequency…..disease is q2…1/1000000

A woman who is a heterozygous carrier of an X-linked recessive disease gene mates with a phenotypically normal male. The disease gene has a penetrance of 80%. On average, what proportion of this couple’s sons will not be affected with the disorder?

Answer is 60%
Explanation: 
The couple’s (heterozygous carrier mother and phenotypically normal male) sons will be affected with the X-linked disorder (Normally) = 50%.
But the question states that the disease is 80% penetrance, then 80% * 50% = 40% therefore 100% - 40% = 60% sons will not be affected with the disease

Two American Black individuals come to you for genetic counseling and are concerned about their risks of having children with Sickle Cell disease. They know that the risk of being a carrier in that population is one in ten. Which ONE of the following most closely approximates their risk for having a child with Sickle Cell disease?

A) 0
B) 1 in 4
C) 1 in 100
D) 1 in 400
E) 1 in 2500

Answer is 1/400 do not know the explanation
ok see„1 in 10 is the carrier frequency, right?
means that 2pq=1/10
but becuz in in autosomal recessive „p is very close to 1, we can say that
2q=1/10, so q=1/20
now„to have the disease„baby must be homozygous„
meaning we need to know qxq( q square)
so putting our values, 
1/20 x1/20 = 1/400
I hope this clarifies, ( I am sorry didnt know how to put the square sign over q :)

i think we can also calculate this directly.. because father and mother can be considered as part of general population… 

chance of father being carrier = 1/10
chance of mother being carrier = 1/10
child with dis.= 1/4
1/10 x 1/10 x 1/4= 1/400 

1. Patients with Hurler’s syndrome (252800) are known to have mutations
at the L-iduronidase locus. The diagnosis of Hurler’s syndrome is most efficiently
made by analyzing a patient’s DNA for
a. A region of DNA that does not encode RNA
b. Alternative forms of the L-iduronidase gene
c. The entire set of genes in one leukocyte
d. A nucleotide substitution in the L-iduronidase gene
e. The position of the L-iduronidase gene on a chromosome

2. Which of the following statements regarding a double-helical molecule
of DNA is true?
a. All hydroxyl groups of pentoses are involved in linkages
b. Bases are perpendicular to the axis
c. Each strand is identical
d. Each strand has parallel, 5′ to 3′ direction
e. Each strand replicates itself

3. A sample of human DNA is subjected to increasing temperature until
the major fraction exhibits optical density changes due to disruption of its
helix (melting or denaturation). A smaller fraction is atypical in that it
requires a much higher temperature for melting. This smaller, atypical fraction
of DNA must contain a higher content of
a. Adenine plus cytosine
b. Cytosine plus guanine
c. Adenine plus thymine
d. Cytosine plus thymine
e. Adenine plus guanine

4. A newborn baby has a sibling with sickle cell anemia (141900) and is at
risk for the disease. The appropriate diagnostic test for sickle cell anemia in
this baby will include
a. DNA amplification
b. Hemoglobin antibodies
c. DNA restriction
d. Red cell counting
e. DNA fingerprinting

5. A polymorphism is best defined as
a. Cosegregation of alleles
b. One phenotype, multiple genotypes
c. Nonrandom allele association
d. One locus, multiple abnormal alleles
e. One locus, multiple normal alleles

Answers:

B,B,B,A,E…. 

11. A farming couple in Northern Michigan consult their physician about
severe skin rashes and ulcers noted over the past year. They also have lost
many cattle over the past year, and claim that their cattle feed changed in
consistency and smell about 1 year ago. Chemical analysis of the feed
shows high concentrations of polychlorinated biphenyls, a fertilizer related
to known carcinogens. The physician sends the chemical to a laboratory for
carcinogen testing, which is performed initially and rapidly by
a. Inoculation of the chemical into nude mice
b. Incubation of mutant bacteria with the chemical to measure the rate of reverse
or “back” mutations
c. Incubation with stimulated white blood cells to measure the impact on DNA
replication
d. Computer modeling based on the structures of related carcinogens
e. Incubation with mammalian cell cultures to measure the rates of malignant
transformation

12. A child presents with severe growth failure, accelerated aging that
causes adult complications such as diabetes and coronary artery disease,
and microcephaly (small head) due to increased nerve cell death. In vitro
assay of labeled thymidine incorporation reveals decreased levels of DNA
synthesis compared to controls, but normal-sized labeled DNA fragments.
The addition of protein extract from normal cells, gently heated to inactivate
DNA polymerase, restores DNA synthesis in the child’s cell extracts to
normal. Which of the enzymes used in DNA replication is likely to be
defective in this child?
a. DNA-directed DNA polymerase
b. Unwinding proteins
c. DNA polymerase I
d. DNA-directed RNA polymerase
e. DNA ligase

13. Patients with hereditary nonpolyposis colon cancer [HNPCC
(114500)] have genes with microsatellite instability, that is, many regions
containing abnormal, small loops of unpaired DNA. This is a result of a
mutation affecting
a. Mismatch repair
b. Chain break repair
c. Base excision repair
d. Depurination repair
e. Nucleotide excision repair

14. If a completely radioactive double-stranded DNA molecule undergoes
two rounds of replication in a solution free of radioactive label, what is the
radioactivity status of the resulting four double-stranded DNA molecules?
a. Half should contain no radioactivity
b. All should contain radioactivity
c. Half should contain radioactivity in both strands
d. One should contain radioactivity in both strands
e. None should contain radioactivity

Answers: B,B,A,A,E……

6. The process that occurs at the 5 position of cytidine and often correlates
with gene inactivation is
a. Gene conversion
b. Sister chromatid exchange
c. Pseudogene
d. Gene rearrangement
e. DNA methylation

7. The average size of a human gene is
a. 1,000 bp
b. 40,000 bp
c. 2 × 106 bp
d. 1.5 × 108 bp
e. 3 × 109 bp

8. Restriction fragment length polymorphism (RFLP) analysis can only be
used to follow the inheritance of a genetic disease if
a. mRNA probes are used in combination with antibodies
b. The disease-causing mutation is at or closely linked to an altered restriction site
c. Proteins of mutated and normal genes migrate differently upon gel electrophoresis
d. Mutations are outside of restriction sites so that cleaving still occurs
e. Restriction fragments remain the same size but their charge changes

9. It is well known that DNA polymerases synthesize DNA only in the 5′
to 3′ direction. Yet, at the replication fork, both strands of parental DNA are
being replicated with the synthesis of new DNA. How is it possible that
while one strand is being synthesized in the 5′ to 3′ direction, the other
strand appears to be synthesized in the 3′ to 5′ direction? This apparent
paradox is explained by
a. 3′ to 5′ DNA repair enzymes
b. 3′ to 5′ DNA polymerase
c. Okazaki fragments
d. Replication and immediate crossover of the leading strand
e. Lack of RNA primer on one of the strands

10. Given that the chromosomes of mammalian cells may be 20 times as
large as those of Escherichia coli, how can replication of mammalian chromosomes
be carried out in just a few minutes?
a. Eukaryotic DNA polymerases are extraordinarily fast compared with prokaryotic
polymerases
b. The higher temperature of mammalian cells allows for an exponentially higher
replication rate
c. Hundreds of replication forks work simultaneously on each piece of chromosomal
DNA
d. A great many different RNA polymerases carry out replication simultaneously
on chromosomal DNA
e. The presence of histones speeds up the rate of chromosomal DNA replication

Answers: E,B,B,C,C…Ok good!

15. Sickle cell anemia (141900) is the clinical manifestation of homozygous
genes for an abnormal hemoglobin molecule. The mutation in the β
chain is known to produce a single amino acid change. The most likely
mechanism for this mutation is
a. Crossing over
b. Two-base insertion
c. Three-base deletion
d. Nondisjunction
e. Single-base substitution (point mutation)

16. Parents bring their newborn daughter to you for consultation about
diagnosis and management. Their first two children, a boy and a girl, have
a complete form of albinism (203100) with pink irides, blond hair, and
pale skin. Which of the following represents your correct advice concerning
the newborn child?
a. A 1/8 risk for albinism and skin cancer from DNA deletions
b. A 1/8 risk for albinism and skin cancer from DNA cross-linkage
c. A 1/4 risk for albinism and skin cancer from DNA point mutations
d. A 1/4 risk for albinism and skin cancer from DNA deletions
e. A 1/4 risk for albinism and skin cancer from DNA cross-linkage

17. A culture of bacteria not resistant to tetracycline develops an infection
from a virus that is derived from the lysis of tetracycline-resistant bacteria.
Most of the bacterial progeny of the original culture is found to have
become resistant to tetracycline. What phenomenon has occurred?
a. Conjugation
b. Colinearity
c. Recombination
d. Transformation
e. Transduction

18. Following ultraviolet damage of DNA in skin
a. A specific excinuclease detects damaged areas
b. Purine dimers are formed
c. Both strands are cleaved
d. Endonuclease removes the strand
e. DNA hydrolysis does not occur

19. Which of the following statements correctly describes eukaryotic
nuclear chromosomal DNA?
a. Each discontinuous piece making up the chromosomes of eukaryotes is about
the same size as each prokaryotic chromosome
b. Unlike bacterial DNA, no histones are associated with it
c. It is not replicated semiconservatively
d. It is a linear and unbranched molecule
e. It is not associated with a specific membranous organelle

20. Xeroderma pigmentosum (278700) is an inherited human skin disease
that causes a variety of phenotypic changes in skin cells exposed to
sunlight. The molecular basis of the disease appears to be
a. Rapid water loss caused by defects in the cell membrane permeability
b. The inactivation of temperature-sensitive transport enzymes in sunlight
c. The induction of a virulent provirus on ultraviolet exposure
d. The inability of the cells to synthesize carotenoid-type compounds
e. A defect in an excision-repair system that removes thymine dimers from DNA

Answers: E,E,E,A,D,E

21. Which of the following statements describes both the spiral structure
of double-stranded DNA and the spiral structure found in certain segments
of protein?
a. They are repeating spiral structures with intervals of pleated sheets
b. They have four alternative units arranged in polymeric chains
c. They are held together by hydrogen bonding
d. They are α-helical
e. They have covalently linked backbones

22. Which of the following descriptions of DNA replication is not common
to the synthesis of both leading and lagging strands?
a. RNA primer is synthesized
b. DNA polymerase III synthesizes DNA
c. Helicase (rep protein) continuously unwinds duplex DNA at the replication
fork during synthesis
d. Nucleoside monophosphates are added in a 5′ to 3′ direction along the growing
DNA chain
e. DNA ligase repeatedly joins the ends of DNA along the growing strand

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elbamaritza - 04/11/08 22:18

23. Which of the following statements describing restriction endonucleases
is true?
a. They always yield overhanging single-stranded ends
b. They recognize methylated DNA sequences
c. They recognize triplet repeats
d. They cleave both strands in duplex DNA
e. They always yield blunt ends

24. DNA fingerprinting is used for paternity testing and forensic identification
of suspects. Which of the following is the most accurate description
of DNA fingerprinting?
a. DNA can be isolated from blood, skin, or sperm and analyzed for variable patterns
of restriction fragments arising from tandemly repeated sequences
(microsatellites)
b. DNA is copied from blood, skin, or sperm RNA using reverse transcriptase and
analyzed for the pattern of complementary DNAs
c. DNA is isolated from blood, skin, or sperm and its fragment size distribution is
analyzed by gel electrophoresis
d. DNA is isolated from blood, skin, or sperm and hybridized with probes from
the HLA locus to visualize HLA gene patterns
e. DNA is isolated from blood, skin, or sperm, centrifuged to separate satellite
DNA fractions, and analyzed by gel electrophoresis

25. The first drug to be effective against AIDS, including the reduction of
maternal-to-child AIDS transmission by 30%, was AIDS drug azidothymidine
(AZT). Which of the following describes its mechanism of action?
a. It inhibits viral protein synthesis
b. It inhibits RNA synthesis
c. It inhibits viral DNA polymerase
d. It stimulates DNA provirus production
e. It inhibits viral reverse transcriptase

Answers: E,E,D,A,E

26. Which of the following enzymes can polymerize deoxyribonucleotides
into DNA?
a. Primase
b. DNA ligase
c. DNA gyrase
d. RNA polymerase III
e. Reverse transcriptase

27. Which of the following statements correctly describes the recombinant
DNA tool known as plasmids?
a. They are found more commonly in viruses than in bacteria
b. They are single-stranded circles
c. They sometimes enhance bacterial susceptibility to antibiotics
d. They sometimes enhance bacterial resistance to antibiotics
e. They are too small to be useful as vectors for the cloning of mammalian DNA
segments

28. Which of the following molecules is found in a nucleoside?
a. A pyrophosphate group
b. A 1′ base linked to a pentose sugar
c. A 5′-phosphate group linked to a pentose sugar
d. A 3′-phosphate group linked to a pentose sugar
e. A terminal triphosphate

29. Which is the most correct sequence of events in gene repair mechanisms
in patients without a mutated repair process?
a. Nicking, excision, replacement, sealing, recognition
b. Sealing, recognition, nicking, excision, replacement
c. Recognition, nicking, excision, replacement, sealing
d. Nicking, sealing, recognition, excision, replacement
e. Nicking, recognition, excision, sealing, replacement

30. Which of the following enzymes can be described as a DNA-dependent
RNA polymerase?
a. DNA ligase
b. Primase
c. DNA polymerase III
d. DNA polymerase I
e. Reverse transcriptase

Answers: E,D,B,C,B

31. Radiation therapy is employed for many cancers, including irradiation
of the central nervous system to destroy lymphoblasts in leukemia. Which
of the following accounts for the destruction of rapidly growing cells?
a. Cross-linking of DNA
b. Demethylation of DNA
c. Cleavage of DNA double strands
d. Disruption of DNA-RNA transcription complexes
e. Disruption of purine rings in DNA

32. Mammalian chromosomes have specialized structures with highly
repetitive DNA at their ends (telomeres). Which aspect of telomeric DNA
replication is different from that of other chromosomal regions?
a. The DNA polymerase uses an RNA primer but does not degrade it
b. The DNA polymerase contains an RNA molecule that serves as template for
DNA synthesis
c. The DNA polymerase must cross-link the 5′ and 3′ termini
d. The DNA polymerase has a σ subunit that facilitates binding to repetitive DNA
e. The DNA polymerase does not use an RNA template or primer

33. Which statement about the “genetic code” is most accurate?
a. Information is stored as sets of dinucleotide repeats called codons
b. The code is degenerate (i.e., more than one codon may exist for a single amino
acid)
c. Information is stored as sets of trinucleotide repeats called codons
d. There are 64 codons, all of which code for amino acids
e. The sequence of codons that make up a gene exhibits an exact linear correspondence
to the sequence of amino acids in the translated protein

34. Sickle cell anemia (141900) is caused by a point mutation in the
hemoglobin gene, resulting in the substitution of a single amino acid in the
β-globin peptides of hemoglobin. This mutation is best detected by which
of the following?
a. Isolation of DNA from red blood cells followed by polymerase chain reaction
(PCR) amplification and restriction enzyme digestion
b. Isolation of DNA from blood leukocytes followed by Southern blot analysis to
detect globin gene exon sizes
c. Isolation of DNA from blood leukocytes followed by DNA sequencing of globin
gene introns
d. Isolation of DNA from blood leukocytes followed by polymerase chain reaction
(PCR) amplification and allele-specific oligonucleotide (ASO) hybridization
e. Western blot analysis of red blood cell extracts

35. The DNA sequence M, shown below, is the sense strand from a coding
region known to be a mutational “hot spot” for a gene. It encodes amino
acids 21 to 25. Given the genetic and amino acid codes CCC = proline (P),
GCC = alanine (A), TTC = phenylalanine (F), and TAG = stop codon,
which of the following sequences is a frame-shift mutation that causes termination
of the encoded protein?
M 5′-CCC-CCT-AGG-TTC-AGG-3′
a. -CCA-CCT-AGG-TTC-AGG 
b.-GCC-CCT-AGG-TTC-AGG
c.-CCA-CCC-TAG-GTT-CAG
d.-CCC-CTA-GGT-TCA-GG—
e. -CCC-CCT-AGG-AGG——

Answers: E,B,B,D,C

38. The hypothetical “stimulin” gene with two exons encoding a protein of
100 amino acids is found to have abnormal expression in cell culture.
Which of the following mutations would produce a 500-bp stimulin
mRNA and a 133–amino acid peptide that reacts with antibodies to stimulin
protein?
a. Splice junction mutation preventing RNA splicing
b. Frame-shift mutation in codon #2
c. Silent point mutation in the third nucleotide of codon #50
d. Nonsense mutation at codon #2
e. Deletion of exon 1

39. In contrast to DNA polymerase, RNA polymerase
a. Fills in the gap between Okazaki fragments
b. Works only in a 5′ to 3′ direction
c. Edits as it synthesizes
d. Synthesizes RNA primer to initiate DNA synthesis
e. Adds nucleoside monophosphates to the growing polynucleotides

40. The removal of introns and subsequent self-splicing of adjacent exons
occurs in some portions of primary ribosomal RNA transcripts. The splicing
of introns in messenger RNA precursors is
a. RNA-catalyzed in the absence of protein
b. Self-splicing
c. Carried out by spliceosomes
d. Controlled by RNA polymerase
e. Regulated by RNA helicase

1-a
2-d
3-c 


41. A promoter site on DNA
a. Transcribes repressor
b. Initiates transcription
c. Codes for RNA polymerase
d. Regulates termination
e. Translates specific proteins

42. The σ factor found in many bacteria is best described as a
a. Subunit of RNA polymerase responsible for the specificity of the initiation of
transcription of RNA from DNA
b. Subunit of DNA polymerase that allows for synthesis in both 5’ to 3’ and 3’ to 5’
directions
c. Subunit of the 50S ribosome that catalyzes peptide bond synthesis
d. Subunit of the 30S ribosome to which mRNA binds
e. Factor that forms the bridge between the 30S and 50S particles constituting the
70S ribosome

43. An immigrant from eastern Europe is rushed into the emergency room
with nausea, vomiting, diarrhea, and abdominal pain. His family indicates
he has eaten wild mushrooms. They have brought a bag of fresh, uncooked
mushrooms from a batch he had not yet prepared. You note the presence of
Amanita phalloides, the death-cap mushroom. A liver biopsy indicates massive
hepatic necrosis. Care is supportive. A major toxin of the death-cap
mushroom is the hepatotoxic octapeptide α-amanitin, which inhibits
a. DNA primase
b. RNA nuclease
c. DNA ligase
d. RNA polymerase
e. RNA/DNA endonuclease

44. The consensus sequence 5′ TATAAAA 3′ found in eukaryotic genes is
quite similar to a consensus sequence observed in prokaryotes. It is important
as the
a. Only site of binding of RNA polymerase III
b. Promoter for all RNA polymerases
c. Termination site for RNA polymerase II
d. Major binding site of RNA polymerase I
e. First site of binding of a transcription factor for RNA polymerase II

45. The so-called caps of RNA molecules
a. Allow tRNA to be processed
b. Occur at the 3′ end of tRNA
c. Are composed of poly A
d. Are unique to eukaryotic mRNA
e. Allow correct translation of prokaryotic mRNA

Answers: B,A,D,E,D


46. In bacterial RNA synthesis, the function of factor ρ is to
a. Bind catabolite repressor to the promoter region
b. Increase the rate of RNA synthesis
c. Eliminate the binding of RNA polymerase to the promoter
d. Participate in the proper termination of transcription
e. Allow proper initiation of transcription

47. Which of the following statements correctly describes the nucleolus of
a mammalian cell?
a. It differs from that found in bacterial cells in that histones are present
b. It may contain hundreds of copies of genes for different types of ribosomal
RNAs
c. It synthesizes 5S ribosomal RNA
d. It synthesizes 60S and 40S ribosomal subunits
e. It synthesizes all ribosomal RNA primary transcripts

48. Which one of the following statements correctly describes the synthesis
of mammalian messenger RNA (mRNA)?
a. Each mRNA often encodes several different proteins
b. Several different genes may produce identical mRNA molecules
c. There is colinearity of the RNA sequence transcribed from a gene and the amino
acid sequence of its encoded protein
d. The RNA sequence transcribed from a gene is virtually identical to the mRNA
that exits from nucleus to cytoplasm
e. Mammalian mRNA undergoes minimal modification during its maturation

49. Studies of the genetic code in bacteria have revealed that
a. Messenger RNA (mRNA) molecules specify only one polypeptide chain
b. Many triplets can be “nonsense” triplets
c. No signal exists to indicate the end of one codon and the beginning of another
d. The nucleotide on the 5′ end of a triplet has the least specificity for an amino
acid
e. Gene sequence and encoded proteins are not colinear

50. Which one of the following binds to specific nucleotide sequences that
are upstream of the start site of transcription?
a. RNA polymerase
b. Primase
c. Helicase
d. Histone protein
e. Restriction endonuclease

Answers: D,B,B,C,A






A gene product thought to be involved in down regulation of fetal Hb expression is being investigated.Samples of primary tissue cultures of fetal, neonatal and adult liver and bone marrow as well as adequate amount of a DNA probe believed to contain the studied gene, are provided.The best method for determining which of tissue culture samples expresses the studied gene is

A.DNA sequencing
B.Northern blot
C.PCR
D.Southern Blot
E.Western Blot 


ANS=B

Northern Blot is the best technique to determine whether a gene is expressed in a particular cell.

DNA sequencing and Southern blot are both incorrect ,since they examine DNA of a cell and DNA of all the cells is identical.

PCR is used to amplify DNA and cannot determine which tissue culture expressed the gene being studied.In this case DNA probe was supplied in adequate amounts,making PCR unnecessary.

1. An infant exhibits amelia (total absence of a limb). Findings of experimental studies in birds and mice are consistent with the possibility that the limb bud AER failed to express which of the following factors? 
A. HOX gene transcription factor 
B. Bone morphogenetic substance 
C. LIM gene transcription factor 
D. WNT gene factor 
E. Fibroblast growth factor 

2. Recent studies support the likelihood that syndactyly caused by failure of apoptosis between the digital rays may result from disruption of expression of which of the following factors? 
A. Bone morphogenetic substance 
B. Fibroblast growth factor 
C. WNT gene factor 
D. Insulin-like growth factor 
E. LIM gene transcription factor 

3. An infant is born with absence of the ulna and radius and malformations of the carpals and hand. Recent studies show that unique limb segment identities are encoded by HOX genes as follows (scapula - HOX9; humerus - HOX9HOX10; ulna,radius - HOX9HOX10HOX11; proximal carpals - HOX9HOX10HOX11HOX12; distal carpals, metacarpals, phalanges - HOX9HOX10HOX11HOX12HOX13). Therefore, disruption of expression of which of the the following HOX gene families would most likely account for this anomaly? 
A. HOX9 
B. HOX10 
C. HOX11 
D. HOX12 
E. HOX13 

4. An infant born with teratogen-induced phocomelia was probably subjected to which of the following teratogenic substances during the indicated sensitive period. 
substance sensitive period 
A.
B.
C.
D.
E.
alcohol 
thalidomide 
phenytoin 
cadmium 
retinoic acid 
4-8 weeks
2-3 weeks
4-8 weeks
2-3 weeks
4-8 weeks 


5. In cases of mirror polydactyly, which of the following factors, produced within the early limb bud, is both expressed by and restricted to the newly formed caudal orthotopic ZPA and the cranial ectopic ZPA? 
A. BMP4 
B. FGF4 
C. SHH 
D. IGF1 

Check Answers 

1. E 
2. A 
3. C 
4. E 
5. C
. An infant exhibits amelia (total absence of a limb). Findings of experimental studies in birds and mice are consistent with the possibility that the limb bud AER failed to express which of the following factors? 
A. HOX gene transcription factor 
B. Bone morphogenetic substance 
C. LIM gene transcription factor 
D. WNT gene factor 
E. Fibroblast growth factor 


Answer is a 

scapula - HOX9
humerus - HOX9HOX10
ulna,radius - HOX9HOX10HOX11
proximal carpals - HOX9HOX10HOX11HOX12
distal carpals, metacarpals, phalanges - HOX9HOX10HOX11HOX12HOX13








A 33-year-old woman comes to the physician for her first prenatal visit. Her last menstrual period was 7 weeks ago. She has had no bleeding or abdominal pain. She has no medical problems and takes no medications. She has no family history of congenital anomalies. Her husband is 55 years old. He is in good health and also has no family history of birth defects. The patient is concerned that her husband’s age may place their fetus at increased risk of a chromosomal anomaly. She wishes to know the paternal age above which amniocentesis or chorionic villus sampling should be considered. Which of the following is the correct response?


A. Above age 30 

B. Above age 35 

C. Above age 40 

D. Above age 45 

E. There is no age cutoff for paternal risk

The correct answer is E. Increasing maternal age leads to an increased risk of chromosomal anomalies in the fetus. These anomalies include trisomy 13, 18, and 21. Advanced maternal age also leads to increased rates of the sex chromosome aneuploidies 47 XXY and 47 XXX. Because of this relationship between advanced maternal age and chromosomal anomalies, many experts suggest amniocentesis or chorionic villus sampling in women who will be 35 years of age or older at the time of their delivery. Paternal age has not been shown to be related to chromosomal anomalies. There is evidence that advanced paternal age is linked to an increased risk of autosomal dominant mutations, which lead to diseases such as neurofibromatosis, achondroplasia, Apert syndrome, and Marfan syndrome. Increasing paternal age also may be associated with X chromosome mutations that are transmitted through carrier daughters to affected grandsons. However, these risks are exceedingly small, and it is currentl y not possible to screen prenatally for all the autosomal dominant or X-linked diseases that advanced paternal age may be associated with. Therefore, unlike with women, in whom the age of 35 is usually given as the cutoff for chromosomal evaluation of the fetus, there is no age cutoff for paternal risk. 

To state that amniocentesis or chorionic villus sampling should be considered for a paternal age above age 30 (choice A), 35 (choice B), 40 (choice C), or 45 (choice D) is incorrect. As explained above, advanced paternal age is associated with autosomal dominant mutations and X-linked mutations. These mutations are very rare, and we are currently unable to screen for all of these mutations prenatally.

Type III collagen is a homo-trimer of a1 (III) chains. If one of the two copies of the COL3A1 gene (which codes for this protein) contains a missense mutation which produces a mutated (but stable) protein, what fraction of type III collagen molecules will contain a mutant subunit and thus be abnormal?

A 1/8 
B 1/4 
C 1/2 
D 3/4 
E 7/8 

The easy way to find out what fraction of molecules will be abnormal is to find out what fraction of molecules will be normal and subtract from one. This is because there are many abnormal combinations (-++, —+, —-, etc.) but only one normal combination: +++. The fraction of molecules that will be normal is the fraction that gets a molecule derived from the normal copy of the gene three times in a row, or 1/2 x 1/2 x 1/2 = 1/8. The fraction of molecules that will be abnormal is thus 1 - 1/8 or 7/8. 


A neonate is found to have an enzymatic deficiency
in the conversion of pyruvate to pyruvate phosphate. The
wild-type sequence includes the following:
Lys -Arg -His -His -Tyr -Leu
AAG-AAG-CAC-CAC-UAC-CUC
The sequence of the mutated enzyme is
Lys -Glu -Ala -Pro -Leu -Pro
AAG-GAA-GCA-CCA-CUA-CCU
What kind of mutation is illustrated by the above
amino acid sequence?
(A) Point mutation
(B) Frame shift mutation resulting in a nucleotide deletion
(C) Chain termination mutation
(D) Frame shift mutation resulting in the addition of a
nucleotide
(E) Splice mutation

D is correct

Shortly after birth, a male infant was hypotonic with lethargy, weak cry, decreased reflexes, and poor suckling reflex. There was no family history of neuromuscular, developmental, genetic, or feeding disorders. All laboratory tests were normal except a fluorescent in situ hybridization (FISH) chromosome analysis. Which gene is most likely affected in this infant?


A. CFTR 
B. DMD 
C. FMR1 
D. HEXA 
E. SNRPN

This is Prader-Willi syndrome

Option E (SNRPN) is correct. This gene encodes a small ribonucleic protein subunit that is imprinted and must be expressed from the paternal chromogranin 15 (Chr 15) because the maternal allele is methylated. The most common cause of Prader-Willi syndrome, suggested by the symptoms in the vignette, is a deletion of this gene as part of a microdeletion of Chr 15.

Assuming Hardy-Weinberg equilibrium for alleles at the CFTR (cystic fibrosis) locus in the U.S. Caucasian population, and given that the mutant allele frequency, q, is 1/50, what fraction of this population are carriers of a CFTR mutation?

A (1/50)2 
B 2/50 
C 1/50 
D 1/100 
E (49/50)2 


B - Correct. 

The normal allele frequency is calculated using the formula p+q=1; so given the mutant allele frequency of q=1/50, p=49/50. For a population in Hardy-Weinberg equilibrium, the heterozygote carrier frequency is 2pq = 2 x 49/50 x 1/50 = 98/2500. The closest answer of those given is 2/50

A woman has a brother affected with a rare autosomal recessive disorder (she herself is not affected). This disorder is 100% penetrant at birth. She undergoes a carrier screening test which detects carriers with 98% sensitivity, and 5% false positive rate; she tests positive. Which of the following is the best estimate of her risk for being a carrier?

A 100% 
B 97.5% 
C 66.7% 
D 50% 
E 33.3%
b-Correct. 

To solve this problem, apply Bayes’ theorem and construct a table as follows: 
Woman is a Carrier Woman is not a Carrier 
prior prob. 2/3 1/3 

conditional prob. 0.98 0.05 

joint prob. 2/3X0.98=0.653 1/3X0.05=0.017 

posterior prob. 0.653/(0.653+0.017)=0.975 0.017/(0.653+0.017)=0,025

The worldwide distribution of b-thalassemia, sickle cell anemia, and glucose-6-phosphate dehydrogenase (G6PD) deficiency coincides with that of:

A lactase persistence. 
B cholera. 
C malaria. 
D influenza. 
E multiple sclerosis.
ccc
- Correct. 

The sickle cell allele is the best understood example of a “balanced polymorphism”, a mutant allele whose high frequency is the result of some fitness benefit to heterozygotes. Sickle cell anemia is most common in parts of the world where malaria has been common in the past. Heterozygotes for the sickle cell mutation are more resistant to malarial infection, apparently because their red cells are more effectively removed from circulation after infection by the malaria parasite. The thalassemias and G6PD deficiency are found in the same geographic distribution as sickle cell anemia, and resistance to malaria in heterozygotes may also be part of the explanation for the high frequency of these alleles. 
The functions which have been identified for the proteins expressed by cellular proto-oncogenes include all of the following except:

A growth factor 
B component of a signal transduction pathway 
C enzyme involved in DNA mismatch repair 
D growth factor receptor 
E transcription factor

C is the answer

You see a previously healthy 3-day-old male infant who is very lethargic, vomiting, and had a seizure. Which of the laboratory tests would be most appropriate to obtain initially to confirm your suspicions?

A plasma phenylalanine and tyrosine levels 
B a 24 hour urine homogentisic acid level 
C serum ammonia level 
D just a plasma phenylalanine level 
E ornithine transcarbamylase gene sequence

C - Correct. 

In OTC deficiency, an X-linked disorder, a deficiency of the enzyme ornithine transcarbamylase affects ammonium metabolism and leads to high level excretion of orotic acid in the urine. Affected males are usually delivered after a full-term gestation and appear normal at birth. They become symptomatic after they start to ingest protein, resulting in the clinical features outlined in the question. 


HbH disease is rarely observed in offspring of African American couples because:

A an overabundance of a2d2 tetramers is causative of HbH disease. 
B the predominant chromosome associated with a-thalassemia in African Americans is (a-). 
C being affected with sickle cell anemia confers relative resistance to the effects of HbH disease. 
D a-thalassemia is only seen in Asian populations. 
E hemoglobin Bart’s is almost always associated with the genotype (a-/a-) in African Americans.

B is the answer

Despite the large number of unique mutations described in the LDL-R gene that cause familial hypercholesterolemia, certain mutations are found in high frequency in specific populations, e.g., the 9.5 kb 3’ deletion is found in 35% of Finnish patients. This is an example of:

A Variable expressivity 
B Allelic heterogeneity 
C Locus heterogeneity 
D Founder effect 
E Balanced polymorphism

D is the answer

Men with an A to G substitution at position 376 in exon V of their glucose-6-phosphate dehydrogenase gene causing a substitution of aspartic acid for asparagine have the type A variant of G6PD and:

A have 10% of G6PD activity 
B have normal or full G6PD activity 
C will pass this on to their sons 
D will develop hemolytic anemia 
E should avoid fava beans and certain drugs

Answer is b
Single mutation does not change enzyme stability

The cellular phenotype of in-frame insertion or deletion mutations in the ligand binding domain of the LDL-R gene is most likely to be:

A Failure to synthesize any LDL-receptor protein 
B Failure to transport LDL-receptor through the endoplasmic reticulum 
C Failure of the LDL-receptor to localize in clathrin-coated pits 
D Failure of the LDL-receptor to bind LDL 
E Failure of the LDL-receptor to be internalized after binding LDL

Answer is dddd
tay sachs an ARdis caused by a deficiency of hexosaminidase a, is iethal in children.in a population of ashkezi jews biood testing shows d frequency of heterozygous to b 0.1.wat is the probability dat the first child of two individual 4rm this population with no family history of d dis will have tay sachs
a.0.25
b.0.11
c.0.0625
d..0025

here 0.1 is probability of selecting an individual that is carrier 
so for two individual 0.1 * 0.1 and
1/4 coz tay sach is AR ds

pregnant woman with one son with sickle cell dis. wants to know the poss. her new baby is gonna be sickler too (this baby is from a different father). What test woul you order?

a.Father caryotype
b. mother caryotype
c.mother Hb electrophoresis
d.Father Hb electrophoresis

mother doesnt need Hb electrop bc we know she is already a carrier, none of them need caryotype bc isnt dx for sicle cell dis.
We do need to know if the father is carrier or not
Good job
Answer d



Q.Restriction fragment length polymorphisms (RFLP ) may be produced by mutations in the sites for restriction endonucleases. For instance a single base change in the site for the nuclease SaII produces the sequence GTGGAC,
which can no longer be recognized by the enzyme. 
The original sequence recognized by SaII was:

1. GTAGAC
2.GCGGAC
3.CTGGAC
4.GTCGAC
5.GTGTAC

Clue is related to Palindromes…Anyone answer with explanation „thanks

No answer of this.

A 20-year-old primigravida experiences a stillbirth at 28 weeks gestation. An autopsy is performed, and the fetus is found to have a hemivertebra, anal atresia, tracheo-esophageal fistula, and renal multicystic dysplasia. The placenta appears normal, and no amnionic bands are seen and no body wall defects are present. Chromosome analysis reveals a 46, XX karyotype. There is no history of any prior birth in this family with such findings. Five subsequent pregnancies in this woman yield normal term births. Which of the following descriptive terms best applies to the abnormalities in this stillborn fetus? 

A Deformation 

B Disruption 

C Sequence 

D Single gene mutation 

E Association

E) CORRECT. The fetus has the VATER association. An association is less specific for determination of subsequent recurrence risk than a defined syndrome

Deformation of a body part occurs from a problem extrinsic to that part. A good example is oligohydramnios producing chest constriction and pulmonary hypoplasia

A disruption is typically a sporadic event, such as occurs with the limb body wall complex, where embryogenesis is disrupted with adhesions forming amnionic bands. The event causing early amnion disruption is not known.

A sequence follows from a single anomaly, as can happen with urethral obstruction sequence leading to hydronephrosis and renal failure.

A 25-year-old primigravida experiences a spontaneous abortion at 13 weeks gestation. She had an uncomplicated pregnancy up to that time. There is no family history of congenital abnormalities. Her nutritional status is normal. Which of the following is the most likely cause for her fetal loss? 

A Placental abruption 

B Umbilical cord torsion 

C Congenital syphilis 

D Trisomy 16 

E Bicornuate uterus

(D) CORRECT. A major cause for first trimester spontaneous abortion is chromosomal anomalies, of which trisomy 16 is the most common

Q 1.) Which karyotype represents an individual with a balanced chromosome complement? 
A) 69, XYY 
B) 45, XX, der(14;21)(q10;q10) 
C) 46, XX, del(4p) 
D) 46, XY, +der(1)t(1;3)(p31;p21)-3 
E) 46, XY, dup(12)(q11.2;q21.2) 

Q 2.) Ned and Stacey are the parents of Mark, a child affected with a fully penetrant, autosomal recessive disorder that is easily diagnosed at birth, and occurs in the population with an incidence of 1/3600. Neither Ned nor Stacey has this disorder themselves. Their next child, Tony, is born without any apparent signs of the disease. Tony grows up and marries Maria, a woman with no known family history of the disorder. The chance that Tony and Maria’s first child will be affected with the same disorder that affects Mark is closest to which of the following numbers? 
A) 1/45
B) 1/120
C) 1/180 
D) 1/240
E) 1/360

Q 3.) Type 1 albinism is an autosomal recessive disorder that results from a mutation in tyrosinase, resulting in the lack of pigmentation. John and Susan, who are married, are both healthy and not affected with type 1 albinism, although John’s father was affected as was Susan’s maternal grandmother. Susan is pregnant. What is the probability that this fetus will be affected with type 1 albinism? 

A) 1/4
B) 1/8 
C) 1/16
D) 1/24
E) 1/64

B) 45, XX, der(14;21)(q10;q10) 
Explanation: This is the karyotype of an individual with a balanced Robertsonian translocation between the long arms of chromosomes 14 and 21. This individual lacks one normal 14 and one normal 21 and has instead a long chromosome composed of the q arms of both chromosomes (note that there are only 45 centromeres - translocation of two acrocentric chromosomes results in the loss of one centromere). As the short arms of acrocentric chromosomes are very small and contain no essential genetic material, their loss does not result in a clinical phenotype. However, the offspring of these balanced individuals are at risk of inheriting missing or extra parts of the involved chromosomes. 



C) 1/180 
Explanation: Three things must occur for Tony and Maria to have an affected child.
1. Tony must be a carrier for the disorder. As Tony doesn’t show any symptoms of the disorder (which is fully penetrant), he does not carry both copies of the mutation. Subsequently, there are three equally likely remaining genotype options for Tony - homozygous normal, heterozygous with the mutation received from his mother, or heterozygous with the mutation received from his father. The overall probability that Tony is a carrier is therefore 2/3.
2. Maria must also be a carrier. Because Maria has no family history of disorder, we assume her risk of being a carrier is equivalent to the carrier frequency in the population. This can be deduced using Hardy-Weinberg equilibrium. The incidence of this disorder in the population is 1/3600. As mentioned in the text (p. 126), a rough estimate of the carrier frequency can be obtained by doubling the square root of the population frequency - ˜ 2 × 1/60 = 1/30. This is the chance that Maria is a carrier.
3. Both Tony and Maria must pass their mutation on to a child. If both Tony and Maria are carriers, the chance they would have an affected child is 1/4.
To obtain the final probability for this problem, multiply the probabilities from all three events - 2/3 × 1/30 × 1/4 = 2/360 = 1/180. 


B) 1/8 
Explanation: Individuals with autosomal recessive disorders have mutations in each copy of the causative gene. For a child of John and Susan to have type 1 albinism, each parent must inherit one copy of the mutation and then pass it to the child. John’s father had type 1 albinism so John inherited one copy of the tyrosinase mutation from him. There is a 1/2 chance he will pass this along to a child. We know Susan’ mother is an obligate carrier for type 1 albinism (she had to inherit one of the two mutant copies from Susan’ grandmother, who was affected.) Subsequently, there is a 1/4 chance Susan will pass along a tyrosinase mutation to a child - a 1/2 chance she received the mutation her mother inherited from Susan’s grandmother multiplied by a 1/2 chance that she would pass along that mutation to a child. So there is a 1/8 total chance the fetus will inherit two copies of the tryrosinase mutation 

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elbamaritza - 04/11/08 22:19

————————————————-
GENETICS

Chromosomes 
• Humans have 22 pairs of homologous chromosomes (autosomes) and 2 sex chromosomes. 
• Homologous means the same chromosomes have the same genes but may have different versions (alleles) of these genes. 
• The X and Y chromosomes (sex chromosomes) are not homologous. 
• Each gene is represented by a pair of alleles 
• A homozygous condition occurs when both alleles are identical. 
• A heterozygous condition occurs when both alleles are different. 
• Chromosomes are numbered according to size, with 1 being the largest and 21 being the smallest (actually 22 is a little larger than 21, but it was numbered before resolution improved enough to notice!). 
• The sex chromosomes are labeled X and Y and are numbered after Chromosome 22; this position does not correlate to size. 
• When autosomes are arranged numerically by size followed by the sex chromosomes, this is called a karyotype. 
• A karyotype is produced when autosomes are arranged numerically by size followed by the sex chromosomes. 
• A karyotype is designated as the total number of chromosomes followed by the sex chromosomes, i.e. 46, XX or 46, XY. 
• Chromosomes are divided by a centromere into “p” arms and “q” arms, with p (think ‘petite’) being the shorter arm and q being the longer arm. 
• The centromere lies between the p and q arms. 
• The location of the centromere defines the chromosome as follows: 
o Metacentric: p and q arms are of approximately equal lengths 
o Submetacentric: the centromere falls nearer to one end of the chromosome than the other 
o Acrocentric: the centromere is near the terminus 
o Telecentric: the centromere is near the terminus 
Modes of Inheritance

Mendelian Inheritance
Autosomal Dominant 
• The allele is expressed in both homozygous and heterozygous conditions. 
• Most alleles segregate randomly during gamete formation. 
• Characterisitcs of transmission, (i.e. inheritance, of a dominant allele) are: 
o affected children have an affected parent 
o males and females are affected in equal numbers 
o each child of an affected parent has a 50% chance of inheriting the abnormal allele. 
o it occurs in every generation

Autosomal Recessive 
• The allele is expressed in the homozygous condition only. 
• Patients are homozygous for the disorder. 
• Characteristics of transmission of a recessive allele are: 
o generally, parents of an affected individual do not express the recessive allele, (i.e. they are heterozygous for the allele.) 
o two heterozygous parents have a 25% chance of having an affected child. 
o affected parents have only affected children. 
o it affects males and females in equal numbers. 
o it may not be present in successive generations.
• Once it is known that two recessive alleles for a particular condition are not present because the individual does not demonstrate the disorder, there is 66% chance the individual is heterozygous for one normal and one abnormal allele, and 33% chance of being homozygous for two normal alleles.
X-linked 
• The X Chromosome carries many important alleles that have no counterpart on the Y Chromosome. 
• Males carry one X Chromosome, with single alleles of each gene, while females have two alleles for each gene. 
• Generally, all alleles on the X Chromosome are expressed in males whereas the situation in females is not as straightforward. 
• In inheritance: 
o males pass X-linked alleles to all daughters, but not to sons. 
o females pass X-linked alleles to 50% of their sons and daughters. 
• X-linked expression can be dominant or recessive, but since males only have one X-chromosome, they express the disorder either way. 
• Expression may be affected by penetrance and expressivity. 
o an individual may possess a combination of alleles that should be expressed, but a proportion of these individuals has no expression - in these cases, penetrance is incomplete or reduced; expression of alleles as expected in all cases is known as ‘complete penetrance’ 
o the expression of alleles has variable presentations - for example café-au-lait spots may occur in different numbers and positions among the same family members - this demonstrates variable expressivity. 

Non-Mendelian Inheritance
Mitochondrial
• Sometimes called ‘maternal inheritance’, 
o mitochondria are cellular organelles containing their own DNA (mtDNA). Unlike nuclear DNA that is inherited from both parents, mitochondria are inherited only through the mother via cytoplasm in the ova. 
o each somatic cell contains mtDNA. 
o some mtDNA may develop independent mutations. 
o cells with only one type of mtDNA are homoplasmic. 
o cells with mutant and non-mutant mtDNAs are heteroplasmic. 
o as the amount of mutant mtDNA increases, a threshold is reached and the cell no longer compensates with the non-mutant mtDNA; a disease occurs.
U
niparental Disomy (UPD) 
• Both homologous chromosomes come from one parent rather than one chromosome from each parent. 
• This usually results from nondisjunction and then loss of the third chromosome (rescue) during meiosis. 
Genomic Imprinting 
• Some alleles are expressed only if they are inherited from a specific parent. The general mechanism of imprinting is methylation to down-regulate expression of a gene. Consequences of imprinting are seen when: 
o one allele is imprinted and the other allele is nonfunctional, (i.e. deleted or mutated) 
o UPDs have both alleles down-regulated; 
o both alleles are imprinted.
Triplet Repeats 
• Di- or trinucleotides become amplified during gametogenesis and amplification occurs preferentially in a specific parent of origin. 
• A threshold number of repeats will result in an abnormal presentation. 
• Amplifications may occur in the 5’ promoter region of the gene, an exon, an intron, or the 3’ region of the gene. 
Mosaicism 
• Different cells have different karyotypes. 
• The total karyotype reflects the individual cell karyotypes, for example, 45,X/46,XX: 
o this, however, does not imply the percentage of cells with each karyotype. 
o the greater the abnormal karyotype percentage, the worse the clinical presentation. 

Common Genetic Disorders
Autosomal dominant

Marfans Syndrome 
• A mutation in the fibrillin gene. 
• Presenting symptoms include tall stature, arachnodactyly, scoliosis, high-risk of dissecting aortic aneurysm, ectopic lentis, and emphysema. 

Beckwith-Weidemann syndrome 
• One of the most common congenital overgrowth syndromes. 
• Congenital findings include macroglossia, omphalocele, gigantism, hemihypertrophy, advanced bone age, visceromegaly, microcephaly, renal medullary dysplasia, facial nevi, and distinctive ear creases. 
• Correct diagnosis is important because patients are at an increase risk for neoplasia. 
• Patients must be monitored for hypoglycemia and hypocalcemia. 

Osteogenesis imperfecta, type I 
• The most common form of osteogenesis imperfecta is caused by a reduction in Type I procollagen. 
• Patients have skeletal osteopenia, fractures that may be present at birth, and blue sclerae throughout life.

Ehlers-Danlos syndrome 
• A mutation in Type V collagen. 
• Patients present with skin hyperextensibility, joint hypermobility, and abnormal wound healing. 
Autosomal recessive

Cystic Fibrosis 
• The most common autosomal recessive disease. 
• Caused by a mutation in the CFTR gene. 
• The mutation leads to an ineffective chloride transporter. 
• Over 900 mutations cause variable expression of disease. 
• Presenting symptoms include meconium ileus, bronchiectasis, pancreatic problems, and pseudomonas pneumonia. 

Phenylketonuria (PKU) 
• A mutation of the phenylalanine hydroxylase gene (PAH) gives rise to classical PKU. 
• Nonclassical forms arise with other mutations in the phenylalanine to tyrosine pathway affecting the level of tetrahydrobiopterin. 
• Phenylalanine diet restriction after birth will avoid serious mental retardation and neurological sequelae. 


Galactosemia 
• The inability to utilize galactose found in milk. 
• Galactose is not converted to glucose by galactose-1-phosphate uridyl transferase and accumulates in tissues. 
• Infants suffer from malnutrition and failure to thrive if a galactose-free diet is not initiated after birth. 
H
urler syndrome 
• The most severe of the mucopolysaccharidoses, this results from a mutation in alpha-L-iduronidase. 
• Heparin and dermatan sulphate cannot be degraded and accumulate in tissues and urine. 
Trisomy - three of the same chromomsome resulting from nondisjunction

Downs Syndrome 
• Trisomy 21 
• The most common chromosomal cause of mental retardation 
• Patients have 3 copies of chromosome 21. 
• It is the most common cause of mental retardation. 
• Patients present with simian crease, brushfield spots, epicanthal folds, and congenital heart problems, and Alzheimer’s disease by age 40. 
• Can also occur from a translocation, but this is less common. 

Edward’s Syndrome 
• Trisomy 18 
• Patients have 3 copies of chromosome 18. 
• Leads to mental retardation, rocker-bottom feat, congenital heart problems, and flexion deformities. 
• Most patients die by age 1 year. 
• Associated with increased maternal age at birth. 

Patau’s Syndrome 
• Trisomy 13 
• Three copies of chromosome 13 
• Present with mental retardation, cleft lip, cleft palate, polydactyly, and congenital heart problems. 
• Patients generally die by age 6 months. 

Klinefelter’s Syndrome 
• Sex Chromosome (47,XXY). 
• One of the most common forms of primary hypogonadism and infertility in males. 
• These patients are genetically male and present with hypogonadism, tall stature, abnormal upper:lower body ratio and gynecomastia. 
Monosomy
• The presence of only one member of a homologous chromosome pair 
• Results from nondisjunction. 

Turner’s Syndrome 
• Sex Chromosome (45,X). 
• The most common cause of primary amenorrhea. 
• Non-disjunction of the sex chromosome during Meiosis. 
• Patient presents with streak gonads, webbed neck, short stature, and poorly developed genitalia. 
• Patients are also at increased risk for coarctation of the aorta, diabetes, autoimmune disorders, and inflammatory bowel disease. 
X-linked


Duchenne Muscular Dystrophy (DMD) 
• An X-linked disorder that affects only males. 
• The mutation leads to the absence of the dystrophin protein needed to keep skeletal muscle cells intact. 
• Patients present with generalized weakness and muscle wasting in hips, pelvic region, thighs, and shoulders. 
• Calves often are enlarged (pseudohypertrophy) 
Becker’s Muscular Dystrophy 
• An X-linked disorder affecting males only. 
• Compared to DMD it has a later onset (2-16 years of age), but the symptoms are similar, though less severe and progress more slowly. 

Hemophilia A 
• X-linked disease leading to a deficiency of Factor VIII, which leads to blood clotting problems. 
• Affects only males and presents with a history of hemorrhage, hematuria and hemarthrosis. 
Hemophilia B 
• X-linked disease leading to a deficiency of Factor IX. 
• Similar to Hemophilia A, but milder. 

Lesch-Nyhan disease 
• A mutation in the HGPRT gene. The enzyme is necessary for the recycling of purine components and the mutation leads to a build-up of purine degradation byproducts. 
• Symptoms begin to present at 3-6 months of age. 
• Most prominent symptom is self-mutilation, along with mental retardation, gout and orange uric acid crystals in the urine. 
Mitochondrial
MELAS - Mitochondria Encephalomyopathy with Lactic Acidosis and Stroke-like episodes. 
• The most common mutation is in tRNALeu. 
• This mutation causes reduced activity of mitochondrial Complexes I and IV that lead to respiratory chain dysfunction. 
• Lactic acidosis occurs in the blood and CSF. 
• Cerebral necrosis and cortical atrophy develop in patients. 
• Presents with visual defects, blindness, hearing loss, seizures, dementia and loss of consciousness. 
• Mean age of onset is 10 years. 

MERRF - Myoclonic Epilepsy and Ragged Red Fibers. 
• The most frequent mutations occur in tRNALys (>80%), tRNASer, and tRNALeu. 
• Lactic acidosis is variable and ragged red fibers in muscle biopsy are a hallmark. 
• Most frequently presents with myoclonus, epilepsy, ataxia, myopathy and hearing loss. 
LHON - Leber Hereditary Optic Neuropathy. 
• One of three mutations in mitochondrial complex I accounts for approximately 95% of cases. 
• Optic nerve degeneration leads to central vision loss in 2-8 weeks. 
• Generally, the first eye is affected followed by the second eye within 6 months. 
• Mean age of onset varies for each mutation, but the 95th percentile for onset for the three most common mutations is age 50 years. 

Triplet Repeats

Fragile X Syndrome 
• This disorder represents both X-linked inheritance and triplet repeat amplification. 
• Disease occurs due to a mutation of the FMRP gene promoter on chromosome X, which leads to an increase in the size of the non-coding 5’ promoter region of the gene. 
• Amplification of CGGs provides additional cytosines for methylation and gene downregulation. 
• Gene product actually prevents disease. 
• Patients present with mental retardation, long face, large ears, dental problems and large testes. 

Huntington’s disease 
• An autosomal dominant disease. 
• Has 100% penetrance, and presents with progressive dementia with adult onset. 
• Amplification of CAG (codes for glutamine) triplet in exon. 
• Mean age of onset ~40 years. 
• Patients usually die within 10-15 years of first symptoms. 
Myotonic dystrophy 
• The most common form of muscular dystrophy affecting adults. 
• It results from amplification of CTG in the 3’ region of the DMPK gene. 
• Three forms exist - classical, mild, and congenital (the most severe form). 
• Muscular weakness is apparent early in the neck muscles. 
• Distal limb muscles are affected and proximal limb muscles remain stronger throughout disease. (Note that with Duchenne and Becker’s muscular dystrophy the proximal muscles are more affected. 
Microdeletion Syndromes
Cri du Chat Syndrome 
• A deletion of the long arm of Chromosome 5. 
• Patient presents with severe mental retardation, microcephaly, ‘cry of the cat’ sounds, hypertelorism, and low-set ears. 
Prader-Willi Syndrome (PWS) 
• The most common microdeletion syndrome 
• The most common form of genetic obesity 
• The first recognized microdeletion syndrome 
• The first recognized imprinting disorder 
• The first recognized uniparental disomy disorder. 
• The multiple modes of inheritance make this an important disorder to know. 
• The gene responsible is SNRPN on Chromosome 15. 
• An understanding of imprinting is important to understanding PWS. 
• Patient presents with hypotonia (pre- and post-natally) followed by hyperphagia. 
• Hypothalamic hypogonadism, short stature, hypopigmentation, small hands and feet, skin picking, behavioral problems. 
Angelman Syndrome 
• This disorder is often studied along with Prader-Willi syndrome and represents a microdeletion disorder and an imprinting disorder. 
• The gene responsible is UBE3A on Chromosome 15. 
• Patients are severely mentally retarded and often mute with inappropriate laughter.


1-AR inheritance…cystic fibrosis….xeroderma pigmentosum…Tay-sachs….sickle cell anemia….PKU(phenyl ketonuria)…hemochromatosis…albinism…many enzyme def

2-AR….Homozygote…25%.. heterozygotes…50%

2-Tay-sachs….ashkenazi jews….hexosaminidase def….AR…homozygote…25%

3-Example if heterozygote for Tay-sachs is 0.1 then first child with Tay-sachs disease is (0.1)(0.1)(0.25) = 0.0025

4-Cystic fibrosis(CF)…most common AR in whites…pancreatic insufficiency.(85%)…deletion of ..phenylalanine…at position..508…..responsible for protein folding…..

5-CF…sterility in male 95%…..inf staph..pseudomonas…cause of death.= pulmonary disease…..abnormal chloride channel protein(CFTR) gene

5-X-linked dominant…female…skipped generation is unusual….hopophosphatemic rickets….fragyl X synd

6-X-linked recessive…male….male are hemizygous…female are heterozygotes(carrier)…,Duchenne….Lesch-nyhan synd(HGPRT..def)…G6PD def….hemophilia A„B….fabry’didease….Hunter synd(iduronate sulfatase def)
.Menkes disease..(gene encoding cu2+ =copper efflux)

7-X-linked recessive….skipped generation are commonly seen because female can show mild(unaffected)

8- Male to male is not seen in X-linked

9-X-linked recessive…hemophilia A….Duchenne.muscular dystrophy

10-X-linked recessive…if mother is carrier…an affected male chance is 25%

11-X-linked recessive…if mother carrier and affected father…affected child is 50%

12-X chromosome inactivation.. is DNA seen as barr body in interphase…random…fixed(it happens in all descendants the same way)…incomplete(tips of long and short arms are not inactivated)

13-X chromosome inactivation…..random….fixed….incomplete

14-XIST…..primary gene causes X inactivation

15-Hemophilia A….mis-sense mutation in the factor VIII …mild disease

16-Nonsense mutation in the factor VIII…sever disease..because it makes truncated mutation

17-Duchenne muscular dystrophy…..gene dystrophin…death..cardirespiratory failure..frame shift mutation..X-linked

18-X-linked dominant…if mother is affected….50% male and female child affected

19-X-linked dominant..if father is affected…..50% female child affected

20-Autosomal dominant…..multiple generations are affected

21-Autosaml recessive…….skipped generations are seen

22-Imprinting….the disease phenotype is very different ,when its transmitted by mother or father

23-Imprinting …..prader-willi synd(father)……angelman synd(mother) both are ch.15…. deletion

24-Prader willi synd …mental retard..hypogonadism…obesity..deletion on long arm ch.15 transmitted by… father

25-Angelman synd….mental retard…seizures ataxia.. deletion… ch.15.. mother

26-Imprinting is based on …..methilation …of …DNA

27-Down’s synd…..trisomy 21….AML is seen with ..newborns….ALL…in older children

28-Mutation in ACTH receptor…cortisol…androgen…ACTH …adrenal cortex all are affected….but not….aldosteron ..that..is ..made ..in zona glomerulosa is independent of ACTH

29-GFR..M…..G lomerulosa…..F asciculata…..R eticularis …… …………..M edula

30-ACTH…….A ldosteron……..C ortisol……….T estosterone(androgen)……H ypertension(epineph)…

31-GFR…M = ACTH

32-Variable expression…different degree of phenotypes or faces of disease…like sever and mild

33-Variable expression….in mitchondria is named…..Heteroplasmy……means..mutation is seen in only some of the mitochondria…..causes sever or …mild

34-Allelic Heteroplasty…different mutations ..in same locus ….



35- Allelic Heteroplasty….like mis-sense …and ..nonsense ..mutation in factor VIII gene

36-Mitochondrial ..inheritance….sperm cell has no mitochondria…so it comes from mother…..both male and female are affected

37-Mitochondrial inheritance…leber hereritary optic neuropathy…optic nerve damage

38-Mtochondria…disorder….often expresses as …neuropathies and …myopathies

39-Brain and muscle are highly dependent on oxidative phosphorylation

40-MELAS….mitochondrial encephalomyopathy…lactic acidosis….stroke like

41-Myoclonic epilepsy….with ….ragged red muscle fibers

42-Mitochondria…….Leber…..MELAS…….Myoclonic epilepsy

43-X-linked…incidence and gene frequency are the same 

44-Duchenne….X-linked….if incidence is„,1/3000…then gene frequency is the same 1/3000

45-…..p + q = 1…..p = dominant allele q = recessive allele heterozygote carrier = 2q

46-Natural selection…African-American population with sickle cell anemia

47-Natural selection…..cystic fibrosis….resistant to typhoid fever

48-Natural selection…Hemochromatosis….in ..iron poor environment

49-Genetic drift….rapid change in gene frequencies in small population

50-Genetic drift…..Ellis van creveld disease…short stature.polydactyly…heart disease

51-Gene flow…exchange of genes among populations….sickle cell anemia ..seen in 1/600 African-American….

52-Ssickle cell anemia…..mutation in …..beta-globolin gene on chromosome 11

53-Consanguity…..increases incidence of …AR inheritance disease

54-Brain ….neuritic palaues…..neurofibrillary tangles….= ..Down synd..(alzheimers)

55-Edwards’ synd….trisomy 18…rocker-bottom feet

56-Patu synd ..trisomy 13…..polydacyly…cleft lip and palate….renal defect

57-Patu…..Edward…..die …before age 2 y ..average for t18 is 2-3m

58-Variable expression…xeroderma pigmentosum…..hempophila A…neurofibromatosis type 1

59-Incomplete penetrance…in some people disease can not be seen…they are Obligate carrier but their father or mother are affected

60-Incomplete penetrance…hereditary hemochromatosis…familial breast cancer…retinoblastoma

61-Delayed aged of onset..huntington…familial breast cancer….familial colon cancer..
Hemochromatosis…….adult polycystic disease

62-Pleiotropy…..single disease… affects ..multiple organs…Marfan„,.osteogenesis imperfecta

63-Locus heterogeneity…..same disease…but …different loci…..ostegenesis imperfecta. Some on…ch.7 and…some on ….ch.17

64-New mutation…….acondroplasia(80% new mutation)…neurofibromatosis type1 (50% new mutation)

65-Anticipation……trinuclotide repeat…mytonic dystrophy…fragile X synd Huntington

66-Anticipation…recent generations have greater severity or earlier age(younger)

67-Barr body…can be seen in those with more than one X….46XX or 47XXY(klinefelter synd)

68-Barr body can not be seen in turner 45X….or 47XYY

69-Fragile X synd…..leading cause of inherited ….mental retardation

70-Fragile X synd…LONG arm of X chromosome elongation…LONG..ears(large)…LONG…testes(marcro-orchidism)…LONG range of joint motion(hypermobile joint)

71-Fragile X synd…..CGG repeat…..in the 5’ untranslated region of FMR1 gene

72- Tumor suppressor gene…Two-hit ..model….mutation of …both copies(two hits) of genes are necessary to promote tumor growth

73-Tumor suppressor gene…is loss of function through deletion or mutation

74-Proto-oncogenes are mutation ..with gain of function….mutation of 1 copy is enough….CDK4….RET genes

75-Two-hit..model..=..retinoblastoma…familial breast cancer…familial colon cancer…melanoma…neurofibromatosis

76-Tumor suppressor mutation….retinoblastoma…mutation of ..RB1..gene…..wilms tumor…ch11

77-Melanoma….Tumor suppressor ( P16) …and….proto-oncogene(CDK4)

78-Tumor suppressor gene….APC gene…ch 5..adenomatous polyposis coli 85% of colon cancers

79-HNPCC….DNA .mismatch repair….hereditary nonpolyposis colorectal cancer..5% of colon cancers

80-Smoking…mutation in P53 gene results….disabled cell repair…lung cancer

81-BRCA 1 ..ch 17…..BRCA 2….ch 13…..both..are DNA repair genes….breast cancer

82-Li-Frauumeni synd….AD inheritance….breast cancer….mutation in …P53 gene

83-DM type 1……mutation in MHC II….young no insulin…heritability..moderate

84-DM type 2….10 times more common…insulin resistant…heritability…high…
Older…..obese

85-DM type 2…young type…Maturity onset diabetes of young …..MODY….AD inherit

86-Alzheimer…>65y…..early onset..= >..A B* E D C F

172Pericentric…includes centromere… A B * C D E F»>….A D C* B E F

173-Paracentric changes are just one side of centomere

174-Pericentric ..changes are both side of centromere

175-AB woman…CD man…child ..is ..ACC…nondisjunction ..in the father during meiosis II

176-Child…is ..ACD…homologous nondisjunction in the father meiosis I

177-Child…is BBC…chromatid sisters nondisjunction…in the mother meiosis II

178-Nondisjunction causes trisomy or monosomy

179-Triploidy….2 sperm ..and fertilize 1 egg

180-Chance of X linked …daughter carrier of a mother carrier is 25%..and 25% for boy affected

181-Becker muscular dystrophy…Duchenne muscular dystrophy…both are X-linked

182-Becker is milder..frame deletion or insertion

183-duchenne…is frame shift deletion or insertion….causes truncated protein ..sever

184-If nondisjunction of Down syn occur during ..mitosis…not meiosis…this type is uncommon and causes mosaicism….a patient with down synd but normal IQ

185-If incidence of cystic fybrosis is 1/2500…then incidence of carrier is 1/25
..q2 = 1/2500…q = 1/50…2q = 1/25

186-Prader-willi synd….abnormal parent-specific methylation pattern

187-Ehler-Danlos..type IV..locus heterogeneity…mutation in type III collagen…fragyl skin..hypermobile joints..ecchymosis ..varicose veins..intestinal bleeding..dislocations

188-Menkes disease…X-linked…copper def…steely hair..arterial torturosity anemia osteoprosis

189-Highly repetitive sequences…in DNA….satellites(20-175 bp)…minisatellites(20-70 bp)…microsatellites(2-4 bp)

190-Microsattelites….is named STRs…simple tandem repeat..amplified by PCR

191-Microsatellites.+ PCR…for ..paternity testing…forensic cases..gene linkage

192-Microsatellite instability…is seen with loss of….mismatch repair…in..HPNCC..and..endometrial cancer

193-A man with tyrosinase negative albinism(ch 11) mates with a woman with tyrosinase positive albinism(ch 15)….the disease is AR inheritance….children are carrier
No of them show albinism…because..they are on different loci and they get just one of them

194-Large segment deletion of DNA…during meiosis crossover…alpha thalasemia(ch16)
Cri-du-chat..(ch5)

195-Mutation in splice sites..affects interon removal from hnRNA..beta thalasemia(ch11)
Gaucher..tay-sachs

196-Digeorge synd…deletions..in ch 22q

197-Blue eyes..is AR..and if rate is 1/100…the carrier incidence is 18/100
..q2 = 1/100..q=1/10..p= 9/10…carrier = 2pq = 2 (0.1) (0.9) = 18/100

198-Disease X-linked dominant…man is affected and woman is unaffected..penetrance is 60%…the chance of affected girl is….because all girls are affected..so its 100%
And penetrance is 60%…..(1) (0.6) = 0.6…..60%

199-A man with bilateral tumor vestibulocochlear..nerve…is neurofibromatosis type 2…ch 22 ..NF2 tumor suppressor gene

200-Ch 5q..=..APC familial colorectal cancer

201-Ch 13q..=..retinoblastoma…Rb gene

202-Ch 17q = neurofibromatosis type 1…NF1 gene

203-Ch 18q = DCC gene for gastric carcinoma…DPC..= pancreatic carcinoma

204-Intrauterine growth retardation..with normal karyotype is confined placental mosaicism

205-Patu..trisomy 13..poldactyly

206-Heinz bodies ..in blood smear….G6PD def X-linked…..trimethoprin/sulfamethoxazole

207-Osteogenesis imperfecta…ch 7

208-Achondroplasia…Huntington..= ch 4

209-SRY gene is located on …Yp…..inhibits DAX1..» sertoli differentiation….secretion of mullerian inhibiting factor…male phenotype

210-A male karyotype 46XX …is ..because of abnormal cross over then X has SRY gene

211-A female karyotype…46XY…is ..because..ch Y has no SRY gene

212-To show linkage between loci….LOD score…LOD >3…shows linkage
LOD < -2..shows no linkage

213-……1 centimorgan(cM) = 1% recombination frequency….

214-……1 centimorga