Question

6) A colorblind female, blood type A married 2 different men. One man (#1) had type AB blood and was colorblind. The other man (#2) had type B blood and normal color vision. Three offspring resulted from the marriages. If possible, determine the father of each of the following offspring. a) female, Type A blood, normal color vision b) male, Type AB blood, colorblind c) female, Type B blood, colorblind In tomatoes, red fruit color is dominant to yellow. Suppose a tomato plant 7) homozygous for red is crossed with one homozygous for yellow. Determine the phenotype and genotype of a) the F1 offspring b) the offspring of a cross of the F1 back to the red parent c) the offspring of the F1 back to the yellow parent

Answer 1

6. let us consider the women's gene is aa and the first husband's gene is ab

the cross gives the possible offsprings as aa, aa, ab, and ab.

all the offsprings should have blood group AB

and the second husband's genes be BB.

This cross gives offsprings as aB, aB, aB, and aB.

here, anyone, a or B can be dominant and also they can both be expressed at the same time.

(a) so the first child female, type A blood and normal color vision is the 2nd husband's child.

(b) the second child is 1st husband's child as the blood group is AB and is colorblind.

(c) the third child is 2nd husband's offspring as the blood group is B and is colorblind.

7. if the red tomatoes RR is crossed to recessive yellow YY

(a) The F1 generation is RRXYY we get RY, RY, RY and RY as the offsprings which are all heterozygous red.

(b) if F1 is crossed back to red parent we get

RYXRR= RR, RR, RY and where we get two true red parents and two hybrid red offsprings.

(c) if F1 is crossed back to yellow parent we get

RYXYY= RY, YY, YY and YY which have three true yellow parents and one hybrid red.

8. let us consider the free earlobe genes are FF and the attached earlobes are AA.

(a) The women can have only one genotype that is AA.

(b) The man can have two genotypes either FF or FA.

(c) if we consider the man having genotype FF the offsprings will have

AAX FF= FA,FA,FA and FA. where all the children have free earlobes and no child has attached earlobes as the trait is recessive.

if the man has FA as the genotype

FAXAA= FA, AA, AA, and AA where 3 children have attached earlobes and one has heterozygous free earlobes.

9. The male tribble if considered to be BB then all the offsprings will be Bb which is black fur.

But when we consider the genotype of the male to be Bb then we have an equal number of chances of having black and white offsprings with genotype Bb and bb. therefore, the genotype of the male, in this case, is Bb.

10. let us consider the father genes to be Ff where f is the recessive mutant allele which can cause disease. Similarly, let the mother's alleles be Mm where m is the recessive mutant allele which can cause disease. So the F1 generation will be FM, fM, Fm, and fm. So 1out of four offspring can develop the disease so the probability is 1/4.