Question

please help with question 3

Given the following diploid progeny classes: Gametes vocv+.ct+ 94 vt .cv .ct y.cv .ctt vf.ct.ct 65 v .cv .ct 780 792 vt .cv .ct+ y.cv .ct 13 vt .cy .ctt 15 1908 What is the uncorrected map units between the following pairs: v-cv, v-ct, and CV -- ct? 14.2, 19.8, 9.1 14 , 0.06, 0.13, 0.19 O 6.4, 18.5, 13.2 93, 268, 191

Answer 1

Answer:

1). 93, 268, 191

2). v, cv, ct

3). 16.1

Explanation:

Hint: Always parental combinations of genotypes are large numbered than the recombinant genotypes.

Hence, the parental (non-recombinant) genotype is v+ cv+ ct+ / v cv ct.

1).

If single crossover occurs between v & cv.

Normal combination: v+ cv+ / v cv

After crossover: v+ cv / v cv+

v+ cv progeny= 89+15= 104

v cv+ progeny = 94+13 = 107

Total this progeny = 211

The recombination frequency between v&cv = (number of recombinants/Total progeny) 100

RF = (211/1908)100 = 11.06%

2).

If single crossover occurs between cv & ct..

Normal combination: cv+ ct+ / cv ct

After crossover: cv+ ct/ cv ct+

cv+ ct progeny= 65+13 = 78

cv ct+ progeny = 60+15 = 75

Total this progeny = 153

The recombination frequency between cv&ct = (number of recombinants/Total progeny) 100

RF = (153/1908)100 = 8.02%

3).

If single crossover occurs between v & ct.

Normal combination: v+ ct+ / v ct

After crossover: v+ ct / v ct+

v+ ct progeny= 89+65= 154

v ct+ progeny = 94+60 = 154

Total this progeny = 308

The recombination frequency between v&ct = (number of recombinants/Total progeny) 100

RF = (308/1908)100 = 16.14%

Recombination frequency (%) = Distance between the genes (mu)

v------11.06-----cv------8.02mu-----ct

Order of genes is “v cv ct”