147 PM OO Verizon 46 instructureuploadss3amazonawscom 3 of 8

1:47 PM OO Verizon 46% instructure-uploads.s3.amazonaws.com 3 of 8 3. In honeybees, sex determination is based on ploidy: females are diploid, developing from fertilized eggs, whereas males are haploid, developing from unfertilized eggs. A set of crosses was done to examine the linkage among three recessive mutations in honeybees: cd (cordovan, ie. dark red, body color), h (hairless), and ch (chartreuse eye color). The following two crosses were done: Phenotypes of drone (male) progeny: Queen genotype cd h 294 cordovan, 236 hairless, 262 cordovan and hairless, 289 wild type cd h h ch 3131 hairless, 3064 chartreuse, 96 chartreuse and hairless, 132 wild type h ch a) Only the genotype of the queen is given. Why is the genotype of the father in each cross not needed for mapping these genes? pt) b) For each cross, determine the genotypes and phenotypes of recombinant and non-recombinant progeny. (4 pts)

Solution

Answer:

3. a. Since haploid males develop from unfertilized eggs, they receive no genes from the male parent. Therefore, the genotype of the male parent does not contribute to the genotype or phenotype of the progeny.

b. For cd and h:

Nonrecombinants: 294 cordovan and 236 hairless

Recombinants: 262 cordovan and hairless, and 289 wild-type.

RF = (262 +289) / (262 + 289 + 294 + 236) = .51 = 51%.

Thus, the two loci are unlinked.

For h and ch:

3131 hairless + 3064 chartreuse = 6195 Non-recombinants

96 chartreuse and hairless + 132 wild-type = 228 Recombinants

RF = 228 / (6195+228) = .035 = 3.5%

c. The map distances corresponds to the recombination frequencies and have been calculated in (b) above. Based upon the RF, we can draw the linkage map.

-------h------------------ch-------- --------cd-------------

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