In Drosophila , the map positions of genes are given in map units numbering from
ID: 254786 • Letter: I
Question
In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u.. long. The X-linked gene for body color-with two alleles, y +for gray body and y for yellow body-resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w +for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f + for normal bristles and f for forked bristles, is located at map position 56.7. Each gene resides on the Xchromosome, and at each locus the wild-type allele is dominant over the mutant allele.
Predict the frequency of gray with white eyes and forked bristles progeny type produced by this mating.
Enter your answer as a percentage to three decimal places (example 0.235 or 0.230).
Frequency=_________
2.
A wild-type trihybrid soybean plant is crossed to a pure-breeding soybean plant with the recessive phenotypes pale leaf l, oval seed (r), and short height (t). The results of the three-point test cross are shown in the table. Traits not listed are wild type.
Calculate the interference value for these data.
Enter your answer to three decimal places (example 0.201).
Phenotype Number Pale 648 Pale, oval 64 Pale, short 10 Pale, oval, short 102 Oval 6 Oval, short 618 Short 84 Wild type 98 1630Explanation / Answer
Answer:
2). Interference: 0.263
Explanation:
Phenotype
Number
Genotype
Pale
648
l R T
Pale, oval
64
l r T
Pale, short
10
l R t
Pale, oval, short
102
l r t
Oval
6
L r T
Oval, short
618
L r t
Short
84
L R t
Wild type
98
L R T
1630
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes are more in number and their trihybrid genotype is L r t / l R T
1).
If single crossover occurs between L & r..
Normal combination: Lr / l R
After crossover: LR/lr
LR progeny= 84+98 = 182
lr progeny = 64+102 = 166
Total this progeny = 348
The recombination frequency between L&r = (number of recombinants/Total progeny) 100
RF = (348/1630)100 = 21.35%
2).
If single crossover occurs between r & t..
Normal combination: rt / RT
After crossover: rT/Rt
rT progeny= 64+6=70
Rt progeny = 84+10=94
Total this progeny = 164
The recombination frequency between r&t = (number of recombinants/Total progeny) 100
RF = (164/1630)100 = 10.06%
3).
If single crossover occurs between L & t..
Normal combination: Lt / lT
After crossover: LT/lt
LT progeny= 6+98= 104
lt progeny = 10+102=112
Total this progeny = 216
The recombination frequency between L&t = (number of recombinants/Total progeny) 100
RF = (216/1630)100 = 13.25%
Recombination frequency (%) = Distance between the genes (cM)
L----------13.25cM--------T-----------10.06 cM--------------R
Expected double crossover frequency = (RF between h & w) * (RF between w & b)
= 13.25% * 10.06% = 0.0133
The observed double crossover frequency = 10+6 / 1630 = 0.0098
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.0098 / 0.0133
= 0.737
Interference = 1-COC
= 1-0.737 = 0.263
1). The parental phenotypes or genotypes are not mentioned in the question. It is not possible to answer
Phenotype
Number
Genotype
Pale
648
l R T
Pale, oval
64
l r T
Pale, short
10
l R t
Pale, oval, short
102
l r t
Oval
6
L r T
Oval, short
618
L r t
Short
84
L R t
Wild type
98
L R T
1630