Answer:
0.035
Explanation:
<u>cv+ is the wild-type dominant allele over cv, therefore:</u>
- cv+cv+ and cv+cv cause wild-type phenotype for crossveinless
- cv cv causes the crossveinless phenotype
<u>Sb is a dominant mutant allele over wild-type Sb+, therefore:</u>
- Sb Sb and Sb Sb+ cause Stubble phenotype
- Sb+ Sb+ causes wild type phenotype for Stubble
It's the cross between the heterozygous female with a homozygous recessive male. Remember that cv and Sb+ are the recessive alleles.
X
-The male produces only 1 type of gamete: cv Sb+
-The female produces 4 types of gametes:
- cv Sb+ ] Parental
- cv+ Sb ] Parental
- cv Sb ] Recombinant
- cv+ Sb+ ] Recombinant
The genes are linked and separated by 7 map units. A distance of 7 mu means that 7% of the resulting gametes will be recombinant. Because there are 2 possible recombinant gametes, each of them will appear in 3.5% of the cases.
The genotypes and proportions of the offspring resulting from the test cross can be seen in the Punnett Square. The phenotypically wild-type individuals will have the genotype cv+ Sb+ / cv Sb+ (heterozygous for crossveinless and homozygous recessive for Stubble) and a 0.035 proportion.
