PRIMERANO PRACTICE PROBLEMS- Leah Flashcards Preview

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Flashcards in PRIMERANO PRACTICE PROBLEMS- Leah Deck (18):
1

WHAT IS THE OFFSPRINGS RISK OF DEVELOPING PKU?

Assume Mom = Type I = PAH homozygous mutant and BH4 synthesis homozygous normal

Assume Dad = Type III = PAH homozygous normal and BH4 synthesis homozygous mutant

Offspring will have NO risk of disease Will be HETEROGENEOUS at both loci

2

PRACTICE PROBLEM:

In a population of 100, there are 36 with MM genotypes, 48 with MN genotypes, and 16 with NN genotypes.

What is the prequency of the M ALLELE? The N ALLELE?

M allele: 2(MM) + MN/ 2(population)

M allele: 2(36) + 48/ 2(100)

=72+48/200= 0.6


N allele: 2(NN) + MN/ 2(population)

N allele: 2(16) + 48/2(100)

=32+ 48/200= 0.4

3

PRACTICE PROBLEM

Hemophilia A (Factor VIII deficiency) is XR.

If the frequency of males with disease in a given population is 1/10,000...

then what is the frequency of female carries of the disease gene?

x=1/10,000

X= 1-x ... therefore 9,999/10,000 ~ 1

Female carriers = 2Xx = 2(1/10k)(1) = 0.0002 ~0.02%

4

PRACTICE PROBLEM

If the frequency of albinos (AR trait, aa) in a population is 1/20,000...

Walk me through finding the frequency of CARRIERS (Aa) and homozygous normal (AA).

a2 = 1/20k = 0.00005; therefore a= 0.007

A=1-a = 0.993 (A+a = 1 per HW LAW)

(0.993)2 = A2= 0.986 AA homozygous frequency

2Aa = (1) (0.007) (2) = 0.014 heterozygote Aa frequency

5

If a set of parents are each carriers of a recessive trait, what is the probability that their offspring will have the recessive disease phenotype?

1/4

(50% chance transmission from each parent.... 1/2 x 1/2 = 1/4)

6

In problem solving, what are the appropriate steps for determining if an offspring will have a disease?

-probability that parent 1 is affected (100%) or a carrier (determine from pedigree or general population frequency)
-probability that parent 2 is affected (1) or a carrier
-probability that offspring is affected (using punnet square)
-multiple these three probabilities

7

Sickle cell disease, an autosomal recessive condition, occurs with a frequency of 1/625 among African Americans.

Patricia has a brother George who has sickle
cell disease and normal parents. Patricia is married to Julius who has no family history of the disease.

What isthe probability that Patricia and Julius will have an affected child?

-Probability that Patricia is a carrier: 2/3
(both of her parents are carriers, so she is either Aa/Aa/AA/aa, we know she is unaffected, so that knocks out aa)

-Probability that George is a carrier = 2pq = 2(1) (1/25)
(q= sqrt 1/625 = 1/25)

-Probability that George and Patricia will have an affected child assuming they are both carriers (1/4)

-Probability of affected child = (2/3) (2/25) (1/4) =4/300= 1/75

8

Jane is a twenty six year old woman brother died of cystic fibrosis. Jane’s parents are both phenotypically normal. Her husband Edward, 30 years of age, is unrelated but has a daughter with CF from a previous marriage (first wife is Rachel). Both
Edward and Rachel are clinically normal.

What is the probability that Jane and
Edward will have a child with CF?

(If Edward had no child by a previous marriage,
what information would you need?)

Probability that Edward is a carrier= 100% (1)
Probability that Jane is a carrier = 2/3
(see previous problem for explanation)
Probability that child will be affected assuming they are both carriers = 1/4

-Probability of affected child = (1) (1/4) (2/3) = 2/12 = 1/6

9

Nathan has an autosomal recessive form of albinism and is married to Nancy who has a brother with albinism. Both of Nancy’s parents are phenotypically normal.


Assume that the same genetic locus is the underlying cause in both sides of the
family. What is the likelihood that Nancy and Nathan will have an affected offspring?

Probability that nathan is aa= 100% or 1.
Probability that Nancy is a carrier= 2/3
Probability that they will have an affected child assuming nancy is a carrier = 1/2

-Probability of affected child = 1 (2/3) (1/2) = 2/6 1/3

10

Ron and his maternal grandfather Barry both have hemophilia A (an X-linked recessive disorder).
Ron marries Diane whose is the daughter of Ron’s maternal aunt.

Assume that Barry’s wife is homozygous normal and that Ron’s aunt’s husband is hemizygous normal and that heterozygous phenotypes will be clinically normal.

1) What is the probability that Ron and Diane will have an affected child? 1)What is the
probability that Ron and Diane will have an affected daughter? 1)What is the probability
of an affected son?

Probability that Ron is a carrier= 100% = 1
Probability that Diane is a carrier= 1/2 (her mom was 100% a carrier)
Probability that they will have an affected CHILD assuming diane is a carrier= 1/2

Probability of "an affected child" = 1 (1/2) (1/2) = 1/4
Probability that they will have an affected FEMALE = 1/8
Probability that they will have an affected MALE =1/8

11

In a population in HW equilibrium, three genotypes are present in the following frequencies:
AA 0.81, Aa 0.18 and aa 0.01.

What are the allele frequencies of A and a?

Frequency of A = sqrt AA= 0.9
Frequency of a= 1-A = 0.1

12

Alport syndrome is a rare autosomal recessive disorder caused by defects in the collagen 4 gene and characterized by nephritis and deafness.

If an Alport woman marries a normal
Danish man with no family history, what is the probability that their child will have Alport
syndrome?

The genotype frequency of Alport syndrome in Danes one is 1/8100. Drawing a pedigree and Punnett square will help.

In your calculation, please round p =1.

Alport woman = 100% carrier= probability 1
Probability that Danish man is a CARRIER (2pq)
p= 1 q = sqrt 1/8100= 1/90
2pq= 2 (1) (1/90) = 2/90

Assuming they are both carriers, probability of a child being affected = 1/2

Probability of affected baby = 1 (2/90) (1/2) = 1/90

13

Phenylketonuria is a rare autosomal recessive disorder caused by the inability to convert
phenylalanine to tyrosine. If newborns with type I
PKU are placed on low phenylalanine diet at
birth, they have normal mental and physical development.

If a patient with PKU marries a normal
man from Scotland with no family history, what is the probability that their child will have PKU?

The frequency of type I PKU homozygotes in the Scottish population = 1/4900. Round p = 1.

Hints: you must determine three probabilities and drawing a pedigree and Punnett square will
help

Probability mom is aa= 100% or 1
Probability that dad is carrier = 2pq or 2(1) (sqrt 1/ 4900) = 2/70 = 1/35
Probability that child will be affected assuming dad is a carrier? 1/2

Probability of an affected child? (1)(1/35)(1/2)= 1/70

14

How do you determine penetrance?

# with phenotype/ # with KNOWN genotype

(For example, go back to pedigree in question five at the end of the practice problems for population genetics). Answer is 60%

15

In an autosomal dominant condition, if an offspring is affected, what must this mean?

-A parent was a carrier or affected.

16

The allele frequency of Huntington's disease (autosomal dominant) is 5 per 100,000 and the
fitness is 99%. What is the mutation rate at the HD locus?

s= 1-f
m= s x q

s = 1-0.99 ~0.01
m= 0.01 (5/100,000)
~0.05/100,000
.....or 5 in every 10,000,000

17

Retinoblastoma segregates as an
autosomal dominant trait.

In a large genotyping study, 200
individuals were shown to carry the defective allele. Of these 200, 20 failed to develop the cancer
at any time in their lives.

Based on this evidence, what is the
percent penetrance of retinoblastoma trait?

penetrance = 180/200 ~90%

18

Tay Sachs is AR.
The population frequency of disease in Ashenkazi Jews is 1/3600. In the general population it is 1/360,000.
What is the CARRIER frequency in each population?

-CARRIER FREQUENCY = 2pq
-ALWAYS ASSUME p=1
-q^2= disease frequency
- SO FOR JEWS..... q = sqrt (1/3600)
-So FOR EVERYONE ELSE q= sqrt (1/360,000)

SO FOR JEWS THE CARRIER FREQUENCY IS:
2pq= 2(1)(1/60) =1/30
FOR THE GENERAL POPULATION:
2pq = 2(1)(1/600)= 1/300

Ten times more carrier in the Ashenkazi Jewish population.