B) You can conclude that A 1 is identical by descent. Notice that A2 had to come from the father (and therefore, A1 is maternal in both cases).

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1 Homework questions. Please provide your answers on a separate sheet. Examine the following pedigree. A 1,2 B 1,2 A 1,3 B 1,3 A 1,2 B 1,2 A 1,2 B 1,3 1. (1 point) The A 1 alleles in the two brothers are identical by state (this just means that they are both A 1 ). Can you infer that they are also identical by descent? B) You can conclude that A 1 is identical by descent. Notice that A2 had to come from the father (and therefore, A1 is maternal in both cases). 2. (1 point) The B 1 alleles in the two brothers are identical by state (this just means that they are both B 1 ). Can you infer that they are also identical by descent? C) You can conclude B 1 is not identical by descent. Notice, for the brother on left, that B2 had to come from the father (and therefore, B1 is maternal). For the brother on the right, B3 is maternal so B1 is paternal. Thus, B1 is maternal in one case and paternal in the other case. (Questions 3-5) Consider two populations, 1 and 2, that differ at two unlinked loci, A and B. In each population a specific allele is fixed at each locus (i.e. all individuals in population 1 are homozygous for A 1 and B 1 -- they have the genotype A 1,1 B 1,1 -- while all individuals in population 2 have the genotype A 2,2 B 2,2 ). You allow a large and equivalent number of individuals from the two populations -- for example, 500 males and 500 females from population 1 and 500 males and 500 females from population 2 -- to mate at random (and they do mate at random). maternal A1, B1 maternal A2, B2 paternal A1, B1 A1/A1 B1/B1 A1/A2 B1/B2 paternal A2, B2 A1/A2 B1/B2 A2/A2 B2/B2 Thus ¼ will be A1/A1 B1/B1, ½ will be A1/A2 B1/B2 and ¼ will be A2/A2 B2/B2. 1

2 3. (1 point) Is locus A at Hardy-Weinberg equilibrium in the "G1" generation? Yes, from the forgoing, you can see that for A, the allele frequencies are p = q = 0.5 and the genotype frequencies 0.25, 0.5 and 0.25 correspond to p 2, 2pq and q (1 point) What is the expected frequency of each of the four possible gametes transmitted from the G1 to the G2. The possible genotypes are A1 B1, A1 B2, A2 B1 and A2 B2 3/8 A1 B1, 1/8 A1 B2, 1/8 A2 B1, 3/8 A2 B2 1/4 of the G1 will be A1/A1 B1/B1; they will generate only A1 B1 gametes (2/8) 1/4 of the G1 will be A2/A2 B2/B2; they will generate only A2 B2 gametes (2/8) 1/2 of the G1 will be A1/A2 B1/B2; they will generate equal numbers of the four possibilities (1/8 for each) Summing those yields the answer above. 5. (1 point) Do the two alleles A1 and B1 show genetic association (linkage disequilibrium) in this G1 generation? Yes! There is clearly and excess of A1 B1 and A2 B2 alleles. 6. Suppose you carry out a mapping cross between two parent strains and P2 as described in lecture 15, and you have 16 F2 individuals in your mapping population. The parent strains and P2 have numerous polymorphic differences between them. You assay 3 RFLP genetic markers (A, B and C) to determine the genotypes of each of the F2 individuals. The data you obtain after PCR, restriction digestion and gel electrophoresis is shown below. a. Write the genotypes for each of the 16 individuals (1-16) for each marker (A, B, and C). For example: 1 or (homozygous) or P2 (homozygous) (0.5 pt) A P2 P2 B P2 P2 P2 P2 C P2 P2 P2 b. What is the observed recombination frequency between A and B based on the data? 11/32 = c. What is the observed recombination frequency between A and C? 5/32 = d. What is the observed recombination frequency between B and C? 8/32 =

3 e. Draw a genetic map for this data showing the order of A, B and C and the map units between them. (You do not need to use a mapping function to correct for unseen double crossovers.) (2 pts total for b, c, d and e) A - ~15 cm. -C- ~25 cm. -B f. Which F2 individual(s), if any, carry a double crossover? (0.5 pt) #15 3

4 II This pedigree shows a family affected by an autosomal dominant genetic disease. Genotypes for three linked markers, A, B and C, are shown I 1 2 II 1 2 III The genotypes are: I-1 A1,2 B1,2 C1,2 I-2 A3,3 B3,3 C3,3 II-1 A1,3 B1,3 C1,3 II-2 A2,2 B2,2 C2,2 III-1 A1,2 B1,2 C1,2 III-2 A3,2 B3,2 C3,2 III-3 A1,2 B3,2 C3,2 III-4 A3,2 B3,2 C1,2 III-5 A3,2 B3,2 C3,2 III-6 A1,2 B1,2 C1,2 7. (1 point) Indicate the phase of alleles in individual II-1 by showing his haplotypes. There are four possibilities. They are a) A1 B1 C1 / A3 B3 C3 8. (1 point) What is the order of these three markers? There are three possibilities: ABC 9. (1 point) Ignoring all of the other loci, calculate a lod score for linkage to of the disease to A with θ = 0 6 * log 10 (2) = 1.8 4

5 10. (1 point) Ignoring all other loci, what value of θ would give the highest lod score for linkage of the disease to B? 1/6 = (1 point) What is the value of that maximal lod score (for linkage of the disease to B at the value of θ that gives the highest possible lod score)? log 10 ((5/6) 5 (1/6)/(1/2) 6 ) = A geneticist succeeds in maintaining three colonies (A, B, and C) of human-mouse hybrid cells. The only human chromosomes retained by the hybrid cells are indicated by plus signs (+) in the table: Human Chromosome Hybrid colony A B C The geneticist tests each of the colonies for the presence of 5 enzymes (q, r, s, t, v) with the following results: q is active only in colony C; r is active in all three colonies; s is active only in B and C; t is active only in B; v is not active in any of the colonies. What can you conclude about the locations of the genes responsible for these enzyme activities? (1 pt) q is on chr. 7, r is on chr. 1, s is on chr. 5, t is on chr. 6 v is not on chromosomes As suggested in the assigned reading, what are the advantages of the FISH physical mapping protocol compared with the method of linkage mapping for the initial determination of the chromosomal location of a gene? (1 pt) (from page 361 in the assigned Hartwell reading): i) all clones can be mapped by FISH, but only clones that detect polymorphisms can be mapped by linkage analysis. ii) Linkage mapping requires the analysis of one locus in relation to another, but FISH does not. iii) FISH requires only a single sample on a microscope slide, but linkage analysis requires genotype information from large set of individuals. 5

6 14. A paper in last week's Nature reports the latest results on characterization of human genetic diversity by DNA sequencing. "A map of human genome variation from population-scale sequencing" The 1000 Genomes Project Consortium Nature 467: Look at the paper and find something relevant to what we have been discussing in class that you consider surprising or interesting. What is it? Quote the one or two sentences that best captures the point that interests you and write between 200 and 400 words on why you find this result interesting or surprising. (3 points) This will be graded by checking that you have indeed found something that is relevant to the material in class and not already well known. 15. Suppose you are examining the genome of a newly discovered organism that has a single linear chromosome. You create a genomic DNA library of 5 BAC clones (designated A through E) that each carry 200 kb of the genomic DNA. You next determine 500 bp of DNA sequence at each end of the 5 cloned inserts using Sanger sequencing and two universal primers that anneal on opposite sides of the cloning site of the BAC vector. (That is, you obtain the 500 bp of the insert sequence that is located at each junction with the BAC vector. These are referred to as "BAC end sequences.") Next, you create STSs out of each of the 10 sequences by designing PCR primers that can amplify these 10 sequences. The table below shows which STSs can be amplified from each of the 5 BACs. BAC clone STSs at ends other STSs A 1, 2 4, 5, 7, 10 B 3, 4 2, 9 C 5, 6 1, 8 D 7, 8 1, 5 E 9, 10 2, 4 a. Diagram a physical map of this region that is consistent with the data in the table. Be sure to indicate the relative order of the BAC clones and show the location of the STSs. (1 pt) CDAEB and , or: BEADC and 3, 9, 2, 4, 10, 7, 5, 1, 8, 6 b. What is the range of possible lengths of your contig (provide a maximum and a minimum length in kb)? (1 pt) kb 6

7 Review questions. These questions are related to the homework questions. However, they do not need to be turned in. (Questions 1-5 compare to homework questions 3-5): Consider two populations, 1 and 2, that differ at two unlinked loci, A and B. In each population a specific allele is fixed at each locus (i.e. all individuals in population 1 are homozygous for A 1 and B 1 -- they have the genotype A 1,1 B 1,1 -- while all individuals in population 2 have the genotype A 2,2 B 2,2 ). First, you cross a single male from population 1 with a single female in population 2. Of course, all of the F1 progeny are heterozygous at both loci, A1,2 B1,2. 1. Is locus A at Hardy-Weinberg equilibrium in the F1 generation (your answer would be the same for locus B)? No! All of the individuals are heterozygous!! 2. What is the expected frequency of each of the four possible gametes transmitted from the F1 to the F2. The possible genotypes are A1 B1, A1 B2, A2 B1 and A2 B2 The four should be in equal abundance. 3. Do the two alleles A1 and B1 show genetic association in this F1 generation? (consider the haplotypes that are transmitted by this F1 generation to the F2 generation). 4. What is the expected frequency of each of the nine possible genotypes in the F2 progeny (assuming random mating among the F1)? The possible genotypes are: A1,1 B1,1 ; A1,1 B1,2 ; A1,1 B2,2 ; 1/16 2/16 1/16 A1,2 B1,1 ; A1,2 B1,2 ; A1,2 B2,2, ; 2/16 4/16 2/16 A2,2 B1,1 ; A2,2 B1,2 and A2,2 B2,2 1/16 2/16 1/16 5. Is locus A at Hardy-Weinberg equilibrium in the F2 generation (your answer would be the same for locus B)? Yes. 7

8 6. Consider question 9 (the lod score for gene A). Now, assume that this disease is only 80% penetrant (meaning that people with the causative genotype are get the disease 80% of the time). What is the lod score for linkage to A with θ = 0 under this revised model? Here you have to calculate the probability of each possible allele/disease combination. allele Disease Under M 1, θ = 0 Under M 0, θ = 0 A1 yes A1 no A3 yes A3 no Z = log 10 [P(data M 1 )/ P(data M 0 )] = log 10 [(0.4) 3 (0.5) 3 / (0.2) 3 (0.3) 3 ] = Consider the pedigree used for questions Ignoring all other loci, what value of θ would give the highest lod score for linkage of the disease to C? What is the value of that maximal lod score (for linkage of the disease to C at the value of θ that gives the highest possible lod score)? log 10 ((2/3) 4 (1/3) 2 /(1/2) 6 ) =