Question

Background Information for Question 1:

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Banshees can be grey or white, have seven, five or three tendrils, and red, orange or yellow eyes. Grey (G) is dominant to white (g). Tendrils form a dominance series where, three (T3) > five (T5) > seven (T7). Banshees homozygous for E'E' have red eyes, those homozygous for EYEY have yellow eyes and heterozygotes (E’EY) have orange eyes. G and T are on the same chromosome while E is on a different chromosome. A true breeding grey banshee with five tendrils and red eyes is mated to a true breeding white banshee with three tendrils and yellow eyes. The F1s are then crossed to a true breeding white banshee with seven tendrils and yellow eyes.

Question 1 A)

What is the inheritance pattern of eye colour? Explain your answer.

Multiple choice, Select one:

a. Red is incomplete dominance to yellow and the heterozygotes have orange eyes.

b. Red is dominant to orange and yellow. Orange is recessive to red but dominant to yellow. Yellow is recessive to both red and orange.

c. We need more information to answer the question.

d. Red is dominant to yellow.

e. Red is co-dominant to yellow, and the heterozygotes have orange eyes.

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Question 1 B)

Write out the parent cross (P-generation) giving the genotype and phenotype of both parents. Correctly showing linked alleles where necessary. (2 marks)

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Question 1 C)

What is the genotype and phenotype of the F₁s? Correctly show linked alleles where necessary. (2 marks)

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Question 1 D)

You had crossed the F₁s to a true-breeding white banshee with seven tendrils and yellow eyes.

a) Write out the genotypes for this cross, correctly showing linked alleles where necessary. (2 marks)

b) What proportion of offspring would be white banshees with five tendrils, if eye colour doesn't matter? Give value as a decimal. Show your work. (1 mark)

c) What proportion of offspring would be grey banshees with three tendrils and orange eyes? Show your work. (3 marks)

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Please show all your work so I can understand. Thank you!

Answer 1

A) Note that there are only 2 alleles, but the heterozygous shows an intermediate phenotype, that means this is an incomplete dominance. The answer is: a. Red is incomplete dominance to yellow and the heterozygotes have orange eyes

B) The first one is: GT5/GT5 ErEr

The second one is: gT3/gT3 EyEy

The cross is: GT5/GT5 ErEr x   gT3/gT3 EyEy

C) Now let us make the punnet square:

	G T5 Er
g T3 Ey	GT5/gT3 ErEy

The phenotype is grey, 3 tendrils and orange eyes

D) a) The F1 genotype is GT5/gT3 ErEy, and the genotype for this new banshee gT7/gT7 EyEy.

Then the cross is: GT5/gT3 ErEy x   gT7/gT7 EyEy

Let us do the square assuming no crossing over:

	G T5 Er	G T5 Ey	g T3 Er	g T3 Ey
g T7 Ey	GT5/gT7 ErEy (grey, 5 tendrils, orange)	GT5/gT7 EyEy (grey, 5 tendrils, yellow)	gT3/gT7 ErEy (white, 3 tendrils, orange)	gT3/gT7 EyEy (white, 3 tendrils, yellow)

b) None, note that since the g allele is linked to T3, then there are not white banshees with 5 tendrils. The answer is 0

c) Again none, since G allele is linked to T5, then there are not grey banshees with 3 tendrils of any eye color. The answer is 0

If you assume crossing over as possible then it will change:

	G T5 Er	G T3 Er	G T5 Ey	G T3 Ey	g T3 Er	g T5 Er	g T3 Ey	g T5 Ey
g T7 Ey	GT5/gT7 ErEy (grey, 5 tendrils, orange)	GT3/gT7 ErEy (grey, 3 tendrils, orange)	GT5/gT7 EyEy (grey, 5 tendrils, yellow)	GT3/gT7 EyEy (grey, 3 tendrils, yellow)	gT3/gT7 ErEy (white, 3 tendrils, orange)	gT5/gT7 ErEy (white, 5 tendrils, orange)	gT3/gT7 EyEy (white, 3 tendrils, yellow)	gT5/gT7 EyEy (white, 5 tendrils, yellow)

In this case we have 2 white banshes with 5 tendrils our of 8, then the answer would be 0.25

And we have 1 grey banshee with 3 tendrils and orange eyes out of 8, then the answer is 0.125