14. A cross was made between females expressing the three x-linked recessive tra
ID: 56442 • Letter: 1
Question
14. A cross was made between females expressing the three x-linked recessive traits, scute bristles (sc), sable body (s), and vermillion eyes (v) and wild-type males. All females were wild type in the F1, while all males expressed all three mutant traits. The cross was carried to the F2 generation and 1000 offspring were counted, with the results shown in the following table. No determination of sex was made in the F2 data. (a) Using proper nomenclature, determine the genotypes of the P1 and F1 parents. (b) Determine the sequence of the three genes and the map distance between them. (c) Are there more or fewer double cross-overs than expected? (d) Calculate the coefficient of coincidence; does this represent positive or negative interference?
*given:
Phenotype: Offspring:
sc s v 314
+ + + 280
+ s v 150
sc + + 156
sc + v 46
+ s + 30
sc s + 10
+ + v 14
Explanation / Answer
The genotype of the F1 males is X(sc s v)/Y
So, all the males will express the mutant phenotype due to the recessive X-linked alleles.
Here, all the F1 females are of wild type. Therefore, they will not be homozygous recessive.
Thus, their genotype might be:
X(sc s v)/X or X/X
Now the P1 parents are said to be female - mutant and male - wild type.
Thus, the P1 male should be: X / Y in genotype
Whereas as the P1 female is a mutant, thus she will be homozygous recessive:
X(sc s v)/X(sc s v)
From the cross of these P1 parents it can be concluded that the genotype of the F1 female is X(sc s v)/X
Thus;
F1 female: X(sc s v)/X
P1 male: X/Y
P1 female: X(sc s v)/X(sc s v)
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