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1. How many chromosomes in a somatic cell of the hybrid? 1n of horse = 32 1n of mule = 31 Fertilization would result in 63 chromosomes in zygote Ans: 63 (b) 2. The likelihood that all chromosomes would migrate in any given gamete cell would be (1/2)63 This would have to occur twice for a gamete cell to find a partner with chromosomes from both species Since this would require two independent events, the likelihood is the product of their independent probabilities: (1/2)63 * (1/2)63 = (1/2)126; Ans: (d) 3. Nondisjunction in the homozygous parent (father) can give rise to an aa gamete. What about heterozygote? Suppose cell undergoes nondisjunction in 2nd meiotic metaphase: a a Both heterozygote (mom) and dad can undergo nondisjunction to yield a trisomic child…Ans: (c) 4. Position can influence expression of genes or transmission (e.g. position effect variegation in Drosophila: bringing a euchromatic gene in the vicinity of heterochromatin can influence its expression; myc gene in Burkitt lymphoma) Linkage group can influence gene expession or transmission; (e.g. abl gene; effects of adjacent segregation in reciprocal translocation heterozygotes). Ability to pair and segregate can also influence expression and transmission; (e.g. inversion heterozygotes). Ans: all of the above; (d) z+ x y Deletion 1 Deletion 2 Deletion 3 Deletion 4 Deletion 5 z Pseudodominance: phenotype of deletion heterozygote is mutant (-) for x and y; wild type (+) for z b − − − + + c + − − + + d − + + + + e − + − + + f + + + − + For example, Deletion 1 “uncovers” genes b, d, and e. d e b c a f I a + + − − − 5. Comparing the overlapping deletions, the order or the genes can be deduced: II III IV V VI debcaf 1 (b,d,e) 2 (b,c) 3 (a,b,c,e) 4 (a,f) a 5 5. The order of genes is debcaf or facbed; Ans: (e) none of the above. 6. The gene closest to band IV would be gene c. Ans: (c) 7. strains 5 6 7 8 1 M 0 0 M 2 0 M L 0 3 L 0 0 L 4 M 0 0 M Expected characteristics: F+ will transfer chromosomal gene markers at LOW frequency (L = low recombinants) only to F− cells Hfr will transfer chromosomal gene markers at HIGH frequency (M = many recombinants) only to F− cells F− cells will yield no recombinants if crossed to themselves, but could show either L or M colonies depending on whether they are crossed to F+ or Hfr strains… Using these criteria: Strain 2 is F− Strain 5 is F− Strain 8 is F− Strain 1 is Hfr Strain 4 is Hfr Strain 6 is Hfr Strain 3 is F+ Strain 7 is F+ 7. false. We can get zero recombinants by crossing F− x F−. 8. Strains 1,4, and 6 are Hfr strains. Ans: (d). Agar Type Str A B C D 1 + + + + − 2 + − + + + 3 + + − + + 4 + + + − + Timings of Samples 0 2.5 5 7.5 10 12.5 15 17.5 20 25 30 35 Agar type 1: will select for d+ exconjugants Agar type 2: will select for a+ exconjugants Agar type 3: will select for b+ exconjugants Agar type 4: will select for c+ exconjugants Number of Colonies of Agar of Type 1 2 3 4 0 0 0 0 0 4 0 0 0 60 0 0 10 132 0 0 60 220 0 0 110 315 0 6 140 370 0 67 165 398 0 104 180 414 0 125 182 416 4 138 185 420 35 140 187 425 40 142 If the data set in the above table is graphed the following time of entry mapping data is obtained: 450 400 350 300 250 Series1 Series2 200 Series3 Series4 150 100 50 0 0 5 10 15 20 25 -50 9. The gene order is ori- a – d – c – b Ans: (c) 10. True. d is closer to c than c is to b. Ans: (a) 30 35 40