Question

Practice questions for BIO 340 (Exam 2) I need help with these questions Please. WILL GIVE GOOD RATING
1. Wild type blue-eyed mary has blue flowers. Two genes control the pathway that makes the blue pigment: The product of gene W turns a white precursor into magenta pigment. The product of gene M turns the magenta pigment into blue pigment. Each gene has a recessive loss-of-function allele: w and m, respectively. A double heterozygote (Ww Mm) is self-pollinated. What proportion of offspring will be magenta? 2. For the following recombination frequencies of 7 gene loci (a, b, c, d, e, f, and g), determine linkage groups, the order of the loci in each linkage group, and the distance between the loci of each group. Loci a and b a and c a and d a ande a and f a and g Recombination Loci Frequency 0.50 b and c 0.50 b and d 0.12b and e 0.50 b and f 0.50 b and g 0.04 Recombination Loci Frequency 0.10 cand d 0.50 cand e 0.18 candf 0.50 candg 0.50 d and e Recombination Loci Frequency 0.50 d and f 0.28 d and g 0.50 eandf 0.50e and g 0.50 fand g Recombination Frequency 0.50 0.08 0.50 0.50 0.50

Answer 1

Following gametes are possible with a given genotype

WM, Wm wM and ww

Product of W gene turns white precursor to magenta pigment

The product of M gene turns magenta pigment into blue pigment.

A progeny having WWmm or Wwmm genotype would produce magenta pigment. Due to the presence of a homozygous recessive mm allele, these progeny (WWmm or Wwmm genotype) would fail to turn the magenta into blue pigment.

A self cross would produce progeny with the following genotype. Out of 16 different genotypes, one progeny would have WWmm and   Wwmm genotype. Therefore, 2/16 = 0.125 % progeny will have magenta color.