Answer:
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.
D. functional group isomerism
Answer:
The morphological features that embryos of various animals share between each other the greater the likelihood they are derived from a common ancestor
Explanation:
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The enzyme becomes denatured, where the shape of the active site changes due to the increased heat. this means that the substrate cannot bind to the active site, therefore the rate of reaction slows down