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# Sample Problems & Essay Questions

Question 1: Your class of 100 people has been typed for a two-allele polymorphism (alleles labeled 1 and 2), and the following 3 genotype counts were obtained:

1,1 30

1,2 50

2,2 20

What are the gene frequencies for alleles 1 and 2 in you class?

Answer: F(1) = 0.55

F(2) = 0.45

Question 2: Suppose that the prevalence of hemophilia A (x-linked recessive) among the females of a population is 1/1,000,000. Based on this figure, what is the gene frequency in this population, and what would the prevalence be among males?

Answer: 0.001, 0.001

Question 3:  Continuing question 2, suppose now that half of the females who have hemophilia A are manifesting heterozygotes. Will this raise or lower the gene frequency estimate? Why?

Answer:  Lowers the estimate because half the females have only one copy of the gene.

Question 4: A woman comes to your office for genetic counseling because her niece has cystic fibrosis (autosomal recessive). She is concerned that she might be a CF carrier. What is the probability that this woman is a heterozygous carrier for a cystic fibrosis mutation?

Answer: 1/2

Question 5:  A man and woman are both affected by an autosomal dominant disorder that is lethal in the embryonic period in homozygotes. The disease has 50% penetrance in heterozygotes. On average, what proportion of their live-born offspring will be affected?

Answer: 2/3 x 1/2 = 1/3

Question 6: A woman who is a known heterozygous carrier of a PKU mutation marries her half second cousin (see pedigree). What is the probability that this couple will produce a child with PKU (autosomal recessive disorder)?

Answer: (1/2)^6 x 1/4 = 1/256

Question 7:  A color-blind (X-linked) male with hemophilia A mates with a normal female. They produce a daughter with Turner syndrome who is not color blind. What is the daughter’s risk of developing hemophilia?

Answer: ~0

Question 8:  A woman is affected with a mitochondrial disease. What proportion of her son’s children will be affected with the disease? What proportion of her daughter’s children will be affected with the disease?

Answer: son’s children: 0%

daughter’s children: 100%

Question 9:  An autosomal dominant disorder is transmitted in each of these 3 pedigrees, and each pedigree member has been typed for a two-allele marker. Based on the genotypes in generation 3, what is the recombination rate between the disease locus and the marker locus?

Answer: 0.10

Question 10:  Continuing question 9, what is the LOD score for the hypothesis that r = 0 in these families?

Answer: LODs = log(16/1) + log (8/1) + log (0) = 1.2 + 0.9 + – infinity = – infinity

Question 11: How has cloning and sequencing the cystic fibrosis gene led to a better understanding of causation, variable expression, and pleiotropy in this disorder?

Answer: Cloning the cystic fibrosis gene led to the demonstration that this gene encodes a chloride ion channel necessary for the exit of chloride ions from specialized epithelial cells. Further study showed that the gene product also influences sodium channel activity in these cells. Mutations in the gene lead to an absence of or reduction in the activity of chloride channels. Consequently, chloride and sodium build up in airway and pancreatic cells. The salt imbalance, and its effect on water balance, help to explain why the airway and pancreas become obstructed with thick, heavy secretions. Pleiotropy is explained by the fact that the chloride defects affect cells in the airway. Pancreas, and sweat glands (in the latter, there is a defect of chloride reabsorption, which produces elevated sweat chloride levels).

Different mutations in the gene can lead to variation in expression (allelic heterogeneity). For example, the DF508 mutation results in a nearly complete lack of ion channels and nearly always produces pancreatic insufficiency when inherited in two copies. Other mutations may only reduce ion channel activity and thus do not necessarily produce pancreatic insufficiency.

Question 12:  Match the pedigree with the most likely mode of inheritance. Note that complicating factors, such as reduced penetrance, may be present. Assume that the gene frequency of the disorder in the general population is very low. These answers may be used more than once.

1. autosomal dominant
2. autosomal recessive
3. X-linked recessive
4. X-linked dominant
5. mitochondrial

Answer: b

Question 13:  Match the pedigree with the most likely mode of inheritance. Note that complicating factors, such as reduced penetrance, may be present. Assume that the gene frequency of the disorder in the general population is very low. These answers may be used more than once.

1. autosomal dominant
2. autosomal recessive
3. X-linked recessive
4. X-linked dominant
5. mitochondrial

Answer: c

Question 14:  Match the pedigree with the most likely mode of inheritance. Note that complicating factors, such as reduced penetrance, may be present. Assume that the gene frequency of the disorder in the general population is very low. These answers may be used more than once.

1. autosomal dominant
2. autosomal recessive
3. X-linked recessive
4. X-linked dominant
5. mitochondrial

Answer: a

Question 15:  Match each karyotype shown below with the appropriate description. The same answer can be used more than once. (Sorry that the images are not that clear)

1. Isochromosome
2. Deletion
3. Reciprocal translocation
4. Robertsonian translocation
5. Inversion

Answer: e

Question 16:  Match each karyotype shown below with the appropriate description. The same answer can be used more than once. (Sorry that the images are not that clear)

1. Isochromosome
2. Deletion
3. Reciprocal translocation
4. Robertsonian translocation
5. Inversion

Answer: d

Question 17:  Match each karyotype shown below with the appropriate description. The same answer can be used more than once. (Sorry that the images are not that clear)

1. Isochromosome
2. Deletion
3. Reciprocal translocation
4. Robertsonian translocation
5. Inversion

Answer: b

# Sample Problems & Essay Questions

Question 1: A population has been screened for mutations in the cystic fibrosis (autosomal recessive) gene using DNA testing. The test shows that 1/100 individuals in this population are heterozygous carriers of a cystic fibrosis (CF) mutation. Based on this figure, what is the expected proportion of affected individuals in this population?

Answer: q = 1/200 p = ~1 2pq = 1/100 2q = 1/100 q^2 + 1/40,000

1/40,000

Question 2: Continuing question 1, if a known CF carrier in this population mates with somebody in the general population, what is the probability that the couple will produce a child affected with cystic fibrosis?

Answer: 1/2 x 1/2 x 1/100 = 1/400

Question 3: Imagine that the known carrier in question 2 mates with his first cousin. What is the probability that this couple will produce a child affected with cystic fibrosis?

Answer: 1/8 x 1/4 = 1/32

Question 4: You (or your significant other) have been identified as the culprit in a paternity suit. DNA testing has been performed to establish whether or not you are the father. Four VNTR loci were tested for you, the mother, and the baby in question. Unfortunately for you, the baby’s alleles and your match for all four loci. The frequencies of these alleles in the general population are .05, .01, .01, and .02. What is the probability that somebody else in the general population could be the father of the baby?

Answer: .05 x .01 x .01 x .02 = 10^-7

Question 5: A woman has had two sons who are affected with Duchenne muscular dystrophy (X-linked recessive). She also has two normal sons, and there is no other family history of the disease. This woman also has muscle weakness in her legs. What is the likely explanation for her muscle weakness?

Answer: manifesting heterozygote

Question 6: Consider two linked loci, labeled A and B. Each locus has two alleles, labeled 1 and 2. The frequencies of the alleles in a population are: A1 = 0.6, A2 = 0.4, B1 = 0.7, B2 = 0.3. If there is linkage equilibrium between these two loci in the population, what is the expected frequency of chromosomes carrying a combination of the A1 and B2 alleles?

Answer: (.6 x .3) = 0.18

Question 7: In the accompanying pedigree, each individual has been assayed for a microsatellite repeat polymorphism that is known to be linked to an autosomal dominant disorder that is being transmitted in this family. The polymorphism has six alleles, labeled 1,2,3,4,5, and 6. The genotypes in this family are shown in the autoradiogram. Based on the genotypes of the offspring in generation III, what is the recombination fraction for the disease locus and the linked microsatellite system?

Answer: 3/8

Question 8: Consider the accompanying pedigrees, in which an autosomal dominant disease is segregating. Based on the genotypes in generation III, what is the LOD score for the hypothesis that the recombination fraction for the disease locus and the marker locus (with 4 alleles, 1,2,3, and 4) is zero?

Answer: ((1/2)^4)/((1/4)^4) = 2^4 = 16

16:1

LOD = 1.2

64:1

LOD = 1.8

~3

Question 9: A somatic cell hybridization panel is shown below. It indicates which clones yielded a positive hybridization signal for a DNA segment from a gene we wish to map. Based on the patterns in the hybridization panel, on which chromosome is the DNA segment located?

Answer:  20

Question 10: The accompanying pedigree shows the mating of two individuals who are both heterozygous carriers of mutations that can cause Wilson Disease, an autosomal recessive disorder of copper metabolism. The parents and their three offspring have each been typed for a closely linked microsatellite polymorphism. Based on the autoradiogram, what is the genotype of individual 5 (i.e., normal homozygote, heterozygote, or affected homozygote)?

Answer: Heterozygote

Question 11: In the family shown below, several females have been diagnosed with breast cancer (the mother in generation I has died). Two of the females have been tested for mutations in the BRCA1 gene on chromosome 17, and both of them have mutations in the gene. Yet a linkage analysis performed on the entire family yields a negative LOD score for linkage between a linked marker (r = .03) and breast cancer in this family. List two possible explanations for this finding. (2-3 sentences should be sufficient.)

Answer: Possible explanations include: (1) recombination between the marker and the disease locus, (2) locus heterogeneity or “nongenetic” causes such that other affected family members do not have BRCA1 mutations (3) incomplete penetrance or delayed age of onset in some family members

Question 12: Describe amniocentesis, chorionic villus sampling, and in vitro fertilization diagnosis, discussing the relative advantages and disadvantages of each procedure.

Answer: Amniocentesis is the withdrawal of a small amount of amniotic fluid from the uterine cavity. It is usually performed at approximately 16 weeks post-LMP and provides fetal cells (amniocytes) for karyotyping, diagnosis of some biochemical disorders, and DNA-based genetic diagnosis. It also provides an alpha-fetoprotein level, which can be used to diagnose neural tube defects. The rate of fetal demise as a consequence of the procedure is approximately 1/200. The primary advantages of the procedure are its accuracy and ability to diagnose a wide variety of conditions. A disadvantage is the fact that it is done relatively late during the pregnancy, and fetal cells often have to be cultured (which takes an additional week or more).

Chorionic villus sampling (CVS) is the removal of a small piece of villus material, which contains fetal cells. It can be done either transabdominally or transcervically and is typically performed at 10-11 weeks post-LMP. An advantage is that it can be done earlier in the pregnancy than amniocentesis. Disadvantages include a higher rate of fetal loss (1-1.3%) and the possibility of misdiagnosis as a result of confined placental mosaicism. There is limited (but not entirely convincing) evidence that CVS performed before 10 weeks can cause limb reduction defects.

In vitro fertilization (IVF) diagnosis is performed on an IVF-generated blastomere at the 8- or 16-cell stage. A single cell is removed (this does not harm the blastomere) and diagnosed using PCR to amplify the DNA. If the diagnosis yields a negative result (i.e., no disease) the blastomere can be implanted. If not, it can be stored or destroyed. The primary advantage is that pregnancy termination is not necessary. Disadvantages include the cost, difficulty of the procedure, and possibility for diagnostic error.

Question 13: The autoradiogram shown here most likely represents variation at which type of system?

1. Restriction fragment length polymorphism
2. Microsatellite repeat polymorphism
3. VNTR polymorphism (minisatellite)
4. Rh blood group polymorphism
5. Each of the above is equally likely

Answer: a

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