Question

(a) Sneeches without fingers or arms are sterile and have never reproduced. However, both fingerless and fingernailless sneeches are able to reproduce. A fingerless sneech (with arms) mates with a sneech that has both arms and fingers but no fingernails. Much to their surprise, half of their kids are wild-type with arms, fingers, and fingernails, while the other half are fingerless like one of the parents. Two fingerless children break social taboo and have kids together. They get a strange collection of children in which the ratios are as follows: 3/16 Wild-type (Arms, fingers and fingernails) 4/16 Armless (No arms, fingers or fingernails) 8/16 Fingerless (Arms, but no fingers or fingernails) 1/16 Fingernailless (Arms & fingers but no fingernails) What is the minimum number of genes that could produce these phenotypes?

(b) Which of the following could explain the unusual ratios in the phenotypes?

Group of answer choices

a. One of the genes is lethal when heterozygous

b. One of the genes is lethal when homozygous

c. One of the genes is epistatic to another

d. It is not possible to get these ratios

e. For 3 genes, this is not an unusual ratio of phenotypes

(c) Can two fingernailless individuals produce a fingerless child without a novel mutation? Why or why not?

Answer 1

Answer- A-Minimum two genes will be enough produce such type of phenotypic ratio as the test cross of dihybrid have 16 coloumn similar to the case.

Answer-B The correct option is B and C those are one gene is lethal when homozygous and one gene is epistatic. Because these unusual genotype and phenotype can be result from these two conditions.

Answer-C Yes the heterozygous may produce fingerless offspring as they will make a homozygous offspring which will be fingerless.