SAMPLE MIDTERM QUESTIONS (Prof. Schoen s lectures) Use the information below to answer the next two questions:

Transcription

1 SAMPLE MIDTERM QUESTIONS (Prof. Schoen s lectures) Use the information below to answer the next two questions: Assume that high blood pressure is inherited as an autosomal dominant trait. You genotype the individuals in your pedigrees with many markers and successfully localize the trait to a region on chromosome 3. A1 and A2 are two codominant alleles of the A marker gene on chromosome 3. Our representative family and their genotypes at the A marker are shown below. Note: Within a given generation, individuals are numbered consecutively from left to right (e.g., I-1, I-2 for the A1/A1 and A2/A2 parents in generation I). 1. (1 point) The individual(s) in the pedigree that has (have) produced informative meioses for mapping of the A marker and high blood pressure locus is (are): a. I-1 and I-2 b. II-3 c. III-2, III-4, and III-5 d. III-1 and III-3 Answer b. In order for an individual to produce informative meioses for mapping the disease and marker loci, it must be doubly heterozygous and produce progeny whose genotypes allow inference of the gamete type that was contributed by the individual in question (e.g., whether it was a parental or recombinant type of gamete). Given the information provided in the pedigree, we see that only II-3 fits these criteria. Note: Genotypes of individuals in generation III provide information for mapping, but

2 it is not because of their meioses (i.e., we are not looking at the progeny of generation III!). 2. (2 points) Calculate a LOD score for linkage between the high blood pressure locus and marker A at a distance of 10 cm. a b c d Answer c. We must inspect the 5 testcross progeny present in generation III and consider the gamete types produced by II-3. Note that parental gametes of II-3 are either A1d or A2D (where D denotes the dominant disease allele), and recombinant gametes are A1D or A2d. Among the 5 testcross progeny, there is one that is the product of a parental gamete of II-3 (individual III-3), whereas the remaining 4 progeny are each the product of recombinant gametes of II-3. Thus, the appropriate calculation of the LOD for 10 cm is: LOD = Log [( )*( )/.25 5 ] = (where Log denotes the base 10 log of the quantity in square brackets). 3. (1 point) In a complementation test the number of complementation groups indicates a. the number of genes required for a specific phenotype. b. the penetrance of the phenotype. c. the number of phenotypes for a gene. d. the number of chromosomes in an organism. e. the quantity of gene product required for a phenotype. Answer a. Use the information below to answer the next two questions: In a population, two different rare genetic diseases (extra finger and an eye disease) are observed. In the pedigree below (taken from a family in the population), a cross represents the occurrence of an extra finger and a black square represents the occurrence of the eye disease:

3 4. (1 point) In the above pedigree, what is the most likely pattern of inheritance for the occurrence of an extra finger? a. autosomal dominant b. autosomal recessive c. X-linked dominant d. X- linked recessive e. Y-linked inheritance. Answer a. 5. (1 point) In the above pedigree, what is the most likely pattern of inheritance for the occurrence of eye disease? a. autosomal dominant b. autosomal recessive c. X-linked dominant d. X- linked recessive e. Y-linked inheritance. Answer d.

4 Use the information below to answer the next two questions: The following three recessive markers are known in lab mice: h, hotfoot; o, obese; and wa, waved. A trihybrid of unknown origin is testcrossed, producing the following offspring: hotfoot, obese, waved 357 hotfoot, obese 74 waved 66 obese 79 wildtype 343 hotfoot, waved 61 obese, waved 11 hotfoot 9 6. (2 points) The map distance between the gene pairs o and h, and h and w are respectively: a. 14 cm, 16 cm b. 16 cm, 14 cm c. 16 cm, 16 cm d. 14 cm, 14 cm Answer c. Note that there are excesses of the triple mutant and triple wild type progeny, suggesting linkage of the three genes and a triple heterozygote parent in which all mutant alleles are on one of the homologs, and all wild type on the other. Looking at the pairs of genes (o, h and h, w individually) we see that there are =160 recombinants for o and h. And there are =160 recombinants for h and w. Since there are 1000 progeny in total, this gives map distances of 16 cm for each pair. 7. (1 point) The linear order of the genes is:

5 a. o,h,w b. o,w,h c. h,w,o d. w,o,h e. Both b. and c. above. Answer a. As inferred from the low frequency of double crossovers and maps distances above. 8. (1 point) In cocker spaniels, black color (B) is dominant over red (b), and solid color (S) is dominant over spotted (s). The genes for these two characters are located on different chromosomes. Suppose that a dog with genotype B/b s/s is mated to a dog with genotype b/b S/s. What fraction of their offspring are expected to be black and spotted? a. 1/16 b. ¼ c. 9/16 d. 3/4. Answer b. 9. (1 point) In a cross between a female with genotype A/a B/b c/c d/d and a male with genotype A/a b/b C/c D/d, what proportion of the progeny will be phenotypically identical to the female parent? (Assume independent assortment of all genes and complete dominance). a. 1/8 b. 3/32 c. 1/64 d. 9/16 Answer b.

6 11. (1 point) In Mendelian genetics, null mutations of different genes may interact to produce deviations from the classic dihybrid ratio. Which of the following is most likely due to interactions between genes in the same linear biosynthetic pathway (i.e., Precursor Intermediate Final product). a. 9:3:3:1 b. 9:7 c. 12:3:4 d. 15:1 Answer b. Interactions between null mutations of two genes in this pathway will produce 9:7 ratio of wildtype:mutant. Use the information below to answer the next two questions: The recessive mutations vg (vestigial wings) and br (brown eyes) identify two autosomal genes on the second chromosome of Drosophila melanogaster. When females heterozygous for these genes were crossed with vestigial winged, brown eyed males, the following classes and numbers of progeny (out of 1000) were obtained: Phenotype _ Number wildtype wings, wildtype eyes 180 vestigial wings, wildtype eyes 317 wildtype wings, brown eyes 308 vestigial wings, brown eyes (1 point) Based upon these results, the map distance between the vg and br genes is estimated to be: a map units. b map units. c map units. d. greater than 50 units because all four classes of offspring were observed.. Answer c. Note that the double heterozygote must be of the type + br / vg +, as the excess types of progeny are not either double wild type or double mutant. 13. (2 points) Which homozygous parent stocks could be used to produce the heterozygous females used in the cross described above? a. wildtype wings, wildtype eyes X vestigial wings, brown eyes b. wildtype wings, brown eyes X vestigial wings, wildtype eyes c. wildtype wings, wildtype eyes X wildtype wings, wildtype eyes d. vestigial wings, brown eyes X vestigial wings, brown eye

7 Answer b. Only this pair of parental stocks could give rise to the double heterozgote parent noted in the above answer. Use the information below to answer the next question: A fruit fly of genotype a+/a b+/b is crossed to another fruit fly of genotype a/a b/b. The progeny of this cross were: Genotype Number of Individuals a+/a b+/b 39 a/a b/b 33 a+/a b/b 10 a/a b+/b (2 point) The Chi-Square associated with the test for independent assortment of the two genes is: a (with 3 degrees of freedom) b (with 1 degrees of freedom) c (with 1 degrees of freedom) d (with 3 degrees of freedom) e. Cannot be calculated from the information given. Answer b or d. As stated, the null hypothesis is independent assortment. By rejecting the null, we might be inclined to accept the alternative, namely that the genes are linked. But rejection of the null could also be due to differential viability. Answer b takes into account the possibility of differential viability of the mutation-bearing gametes, whereas answer d does not. However, I noted that there has been confusion over this due to inconsistency in the way in which the text and MegaManual solves these types of problems. Technically, the method that takes viability into account is more accurate, but I will also accept the method that uses a 1:1:1:1 ratio (no viability differences) as the null hypothesis. 15. (1 point) Non-homologous chromosomes: a. are genetically identical. b. are similar, but there are a few minor genetic differences between them. c. pair with each other during meiosis I. d. are found in mitochondria.. Answer e.

8 16. (1 point) An individual who is heterozygous has brittle bone disease disease (in the mutant heterozygote an abnormal copy of the protein wraps around one or two copies of the other and distorts the conformation of the functional trimeric molecule). This type of mutation is called a: a. a leaky mutation b. a dominant-negative mutation c. recessive mutation d. haplo-insufficient mutation e. haplo-sufficient mutation Answer b. Note: Haplo-insufficient mutation is an incorrect answer. While this is a dominant mutation, the phenotypic basis does NOT arise because two functional doses of an enzyme are required for completion of a biosynthetic pathway. Rather, the mechanism is that of poisoning of one protein unit by another (a dominant negative), as discussed in the text and class (and contrasted with haplo-insufficient mutations) under the topic of dominant mutations (see Lecture 8). 17. (1 point) A man with blood type "B" marries a woman with blood type "B". Which of the following blood types might you expect to see in their children? a. Type A and B would be possible. b. Only type O would be possible. c. Only type B would be possible. d. Types B and O would be possible. e. Only Type AB would be possible Answer d.