In drosophila, the genes crossveinless-c and stubble are linked, about 7 map units apart on chromosome 3. cv-c is a recessive mutant allele of crossveinless-c (cv-c+ is wild type), while sb is a dominant mutant allele of stubble (sb+ is wild type). a dihybrid female drosophila with genotype cv-c sb+/cv-c+sb is testcrossed. the proportion of phenotypically wild-type individuals in the progeny of the testcross will be:
a. 0.035.
b. 0.070.
c. 0.350.
d. 0.465
e. 0.930

A. 0.035.

Explanation:

Female : cv-cSb+/cv-c+Sb

Male : cv-cSb/cv-cSb ( since test cross involves mating with a recessive parent )

Their progeny:

cv-cSb+/cv-cSb : parental

cv-c+Sb/cv-cSb : parental

cv-c+Sb+/cv-cSbc : recombinant

cv-cSb/cv-cSb : recombinant

Wild type phenotype (cv-c+Sb+/cv-cSbc) is a recombinant. Distance between the two genes is 7mu. Since map distance is equal to recombination percentage, there will be 7% recombinants in the progeny. Out of total 7% recombinants, 3.5% will be cv-c+Sb+/cv-cSbc( wild type) and rest 3.5% will be cv-cSb/cv-cSb (mutant type).

Hence, proportion of phenotypically wild-type individuals in the progeny of the test cross will be 3.5% or 0.035.

earthquakes in florida are rare because the state is not located near any tectonic plates. in fact, florida is tied for first place as having the fewest earthquakes of any state. much of florida is limestone and seafloor materials that is not conducive to earthquakes.

explanation:

the answer would be   10%