Question

In the Chinese primrose, the alleles A and a result in blue-colored versus slate-colored flowers, B and b in green vs. red stigma, and C and c in short vs. long style. All 3 genes involved are on the same chromosome. A pure-breeding blue-colored, green-stigma, short-style strain was crossed to a pure-breeding slate-colored, red-stigma, long-style strain. The heterozygous F1 strain was testcrossed to a triply homozygous recessive strain. The progeny are as follows: Flower Stigma Style Number of progeny blue green long 14 blue green short 224 slate green long 142 slate green short 118 slate red long 226 blue red long 122 slate red short 16 blue red short 138

The map distance between the genes A/a and B/b is?

Answer 1

A true breeding parents are homozygous for each trait, which means their genotype are SRL/SRL and srl/srl. Therefore, F1 is SRL/srl. The two parental chromosomes have the highest numbers and are SRL 960 and srl 977. Slate green short sRL (27) and blue red long Srl (27). RS recombinants are Rsl and rSL - Green slate long (427) and red blue short (402). Also, percentage recombination = (no. recombinants/total no. progeny) x 100% .

= ((402 + 427 + 27 +27)/3000) * 100% = 29.4%.

The value for green blue long (95) and red slate short (85)

% recombination here is ((85 + 95 + 27 +27)/3000) * 100% = 7.8%

% recombination = ((402 + 427 + 85 + 95)/3000) * 100% = 33.6%

As % recombination is equivalent to the distance between genes in map units, the map can be drawn as:

R------------29.4------S---7.8--L
----------33.6-----------------------