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 B/b and C/c is
a) 7
b) 16
c) 30
d) 31

Answer 1

The map distance between the genes B/b and C/c is approximately 31 cM (centimorgans) or 34.8716% in the Chinese primrose.

Step-by-step explanation:

The map distance between the genes B/b and C/c in the Chinese primrose can be determined based on the progeny obtained from the testcross. The progeny counts are as follows:

FlowerStigmaStyleNumber of Progenybluegreenlong14bluegreenshort224slategreenlong142slategreenshort118slateredlong226blueredlong122slateredshort16blueredshort138

To determine the map distance, we need to calculate the recombination frequencies between the different gene combinations. The recombination frequencies can then be used to calculate the map distance.

Based on the progeny counts, the recombination frequencies are as follows:

• blue-green: 224/876 = 0.2555
• green-long: 142/876 = 0.1621
• blue-long: 14/876 = 0.016%

To calculate the map distance between B/b and C/c, we add the recombination frequencies:

0.2555 + 0.1621 + 0.016% = 34.8716%

Therefore, the map distance between the genes B/b and C/c is approximately 31 cM (centimorgans) or 34.8716%.