2. Consider a 3-point cross in which the fully heterozygous kkII mm was crossed
ID: 3166017 • Letter: 2
Question
2. Consider a 3-point cross in which the fully heterozygous kkII mm was crossed to a fully homozygous recessive kk l1 mm. The table below shows the number of offspring having each of the eight possible genotypes. Construct a genetic map for these loci (5 points), and calculate the interference value (3 points). (Please note that gene k has two alleles, k and k so kk is a heterozygous genotype, etc.) kk II mm* 621 kk ll mm 109 3 kk II+ mm 57 7 608 64 kk ll mm 103 u sagessive to 4 and h is recessive to B. An organismExplanation / Answer
Answer:
Hint:
Parental genotypes are more than any type of recombinant progeny. Hence parental genotype is k+l+m+ / klm
1).
If single crossover occurs between k+ & l+..
Normal combination: k+l+ / kl
After crossover: k+l / kl+
k+l progeny= 64+109=173
kl+ progeny = 103+57=160
Total of this progeny = 333
The recombination frequency between k+&l+ = (number of recombinants/Total progeny) 100
RF = (333/1572)100 = 21.18%
2).
If single crossover occurs between l+ & m+..
Normal combination: l+m+/lm
After crossover: l+m / lm+
l+ m progeny= 3+57=60
f + progeny = 7+64=71
Total this progeny = 131
The recombination frequency between l+&m+ = (number of recombinants/Total progeny) 100
RF = (131/1572)100 = 8.33%
3).
If single crossover occurs between k+ & m+..
Normal combination: k+m+ / km
After crossover: k+m/km+
k+m progeny= 3+109 = 112
km+ progeny = 7+103= 110
Total this progeny =222
The recombination frequency between k+&m+ = (number of recombinants/Total progeny) 100
RF = (222/1572)100 = 14.12%
Recombination frequency (%) = Distance between the genes (cM)
k+----------14.12cM--------m+-----------8.33cM--------------l+
Expected double crossover frequency = (RF between h & w) * (RF between w & b)
= 0.1412 * 0.0833 = 0.0118
The observed double crossover frequency = 7+3 / 1572 = 0.0064
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.0064/0.0118
= 0.54
Interference = 1-COC
= 1-0.54 = 0.46
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