KAPLAN_USMLE_STEP_1_LECTURE_NOTES_2018_BIOCHEMISTRY_and_GENETICS
.pdf
Part II ● Medical Genetics
Answers
1.Answer: A. The female II-1 in this family is heterozygous for the marker (from the gel) and also has an unaffected father. Her mother is a carrier and the bottom band in the mother’s pattern is associated with the diseaseproducing allele of the factor VIII gene. All observations are consistent with II-1 being heterozygous (Xx) for the factor VIII gene. She has no symptoms, so she is not a manifesting heterozygote (choice E). She cannot be homozygous for the disease-producing allele (choice B) because her father is unaffected. Homozygosity for the normal allele (choice C) is inconsistent with the results shown on the gel. She has inherited the chromosome from her mother (bottom band) that carries the mutant factor VIII allele, but from her father she has received a chromosome carrying the normal allele. Note that her father is not affected, and the bottom band in his pattern is in linkage phase with the normal allele of the gene. This is a case where linkage phase is different in the mother and the father. Incomplete penetrance (choice D) is not a good choice because the female (II-1) does not have the disease-producing genotype. She is heterozygous for the recessive and (dominant) normal allele. One would expect from her genotype that she would be unaffected.
2.Answer: C. The blot shows the top band in the patterns of I-1 and II-2 (the proband) is associated with the disease-producing allele. Because the fetus has inherited this marker allele from the mother (II-2) and Marfan disease is dominant, the fetus will develop Marfan disease. Choices A and B are recurrence risks associated with the pedigree data. With no blot to examine, choice B, 50% risk would be correct. Choice D would be correct if the blot from the fetal DNA showed both the bottom band (must be from mother) and the top band (from the unaffected father). Choice E is incorrect because Marfan is a dominant disease with no “carrier” status.
3.Answer: E. Although II-3 has an RFLP pattern consistent with heterozygosity for the PKU allele, she has PKU. The best explanation offered is that recombination has occurred, and although she is heterozygous for the restriction site generating the RFLP pattern, she is homozygous for the mutation causing PKU. The restriction site is 10 million bp upstream from the phenylalanine hydroxylase gene so there is a minimum chance of recombination of 10%. Although this is small, it is the most likely of the options listed. The phenylalanine hydroxylase gene is not on the X chromosome (choice A). Heteroplasmy (choice B) is associated with mitochondrial pedigrees, and the phenylalanine hydroxylase gene is a nuclear one. The RFLP pattern is quite consistent with I-2 being the biologic father (choice C), and he is a known carrier of the PKU mutation because he has another affected child (II-1). If II-3’s RFLP pattern showed homozygosity for the marker (identical to II-1), and she had no symptoms, incomplete penetrance (choice D) would be a good choice.
404
Chapter 6 ● Genetic Diagnosis
4.Answer: C. The disease-producing allele of the gene is associated with the presence of the HindII site. Notice that both affected males show two smaller bands (75 and 40 bp). II-3, a carrier female, also has these two smaller bands in her pattern, in addition to a larger PCR product (115 bp), representing the absence of the HindII site on her normal chromosome. III-2 has only the larger PCR product (notice the density because both chromosomes yielded this product). She is homozygous for the normal allele. Choice A, carrier, would be correct if her pattern had looked like those of II-3 and III-1. All the males shown are hemizygous (choice B) for the dystrophin gene because they have only one copy. II-1 and III-3 are hemizygous for the disease-producing allele, and II-2 is hemizygous for the normal allele. No one in the family is homozygous for the diseaseproducing allele (choice D). In an X-linked pattern, this would be characteristic of a female with two copies of the disease-producing allele and is very rarely seen. III-2 is not a manifesting heterozygote (choice E) because she has no symptoms and is not a heterozygote.
5.Answer: E. The blot indicates that both parents are heterozygous for the mutant allele. Because both are phenotypically normal, the disease must be autosomal recessive. If it had been X-linked recessive, the man would be hemizygous. Thus, the chance they will have an affected child is 25% (0.25).
6.Answer: A. The affected grandfather has marker alleles DS2 and DS3. There is no information about which one is in linkage phase with his diseaseproducing huntingtin allele. On the basis of the pedigree alone, the daughter has a 25% change of inheriting the grandfather’s disease-producing huntingtin allele (choice B); however, she would like more information. Because her father (II-1) does not wish to be tested or know about his genetic status with respect to Huntington’s, it is unethical to test the daughter for the triplet repeat expansion. The results would necessarily reveal the status of her father also. By doing an indirect genetic test, one can see the daughter has inherited one of her marker alleles (DS2) from the grandfather via her father. This means she has a 50% chance of developing Huntington’s because there is a 50% chance that DS2 is a marker for the disease-producing huntingtin allele in the grandfather and a 50% chance it is not (and DS3 is). Notice the result does not reveal additional information about her father (II-1). Before her testing, he had a 50% chance of having the disease-producing huntingtin allele. His risk is still 50% with the information from the daughter’s test. However, if the father (II-1) does develop Huntington’s in the future, that will mean that the daughter has a 100% chance of having the disease also (choice D).
If her marker status had been DS1/DS1, her chances of developing Huntington’s would have been near 0 (choice E) because she did not inherit these alleles from her grandfather. One came from her grandmother (via her father) and one from her mother. This result still would not reveal additional relevant information about her father (II-1), whose risk would remain 50%.
405
