Question

Waxy endosperm (wx), shrunken endosperm (sh), and yellow seedling (v) are encoded by three recessive genes in com that are linked on chromosome 5. A com plant homozygous for all three recessive alleles is crossed with a plant homozygous for all the dominant alleles. The resulting Ft are then crossed with a plant homozygous for the recessive genes in a three-point testcross. The progeny of the testcross are given below:

wx sh V 87
Wx Sh v 94
Wx Sh V 3,479
wx sh v 3,478
Wx sh V 1,515
wx Sh v 1,531
wx Sh V 292
Wx sh v 280
Total 10,756

a. Determine the order of these genes on the chromosome.
b. Calculate the map distance between the genes.

Answer 1

(a) See the answer in the explanation below

(b) Distance between wx and v = 7 mu

Distance between gene v and sh= 30 m.u.

Step-by-step explanation:

(a) The order of the genes are given below;

wx sh V 87 (double crossover)

Wx Sh v 95 (double crossover)

Wx Sh V 3479 (NCO/parental)

wx sh v 3478 (NCO/parental)

Wx sh V 1515 (Single crossover between v and sh)

wx Sh v 1531 (Single crossover between v and sh)

wx Sh V 292 (Single crossover between wx and v)

Wx sh v 280 (Single crossover between wx and v)

Total 10,756

(b) To calculate the distance between genes, we use the formula;

Recombination frequency = number of recombinant progeny/total number of progeny *100

Distance between V and Sh = 1515+1531+87+94/10756 *100

= 3227/10756 *100

= 30% or 30 mu

Therefore, distance between V and Sh = 30 mu

Distance between wx and v = 292+280+87+94/10756 *100

= 753/10756 *100

= 7% or 7 mu

Therefore, distance between wx and v = 7 mu