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.

Answer 1

Q1A) We are told that eye color is determined by two alleles, E^r and E^y, with homozygous combination of E^r leading to red eyes and that of E^y leading to yellow eye color. We are told that the heterozygotes have neither red nor yellow eyes, but orange eyes, suggesting that E^r and E^y are co-dominant and can contribute equally to eye color. Therefore e. Red is co-dominant to yellow, and the heterozygotes have orange eyes is the right answer.

Q1B) Note that G and T are on the same chromosome and are linked alleles. Linked alleles do not segregate independently and are hence represented by an over-bar.

Parental (P) genotype GG T5 T5 E'E gg T3 T3 EYEY Х phenotype Grey banshee with 5 tendrils and red eyes; homozygous at all loci White banshee with 3 tendrils and yellow eyes; homozygous at all loci

Q1C) We know T³ is dominant to T⁵, and G is dominant to g, so the heterozygotes will have 3 tendrils and grey color. However, E^y and E^r are co-dominant, and hence the F1 heterozygotes will have orange eyes. This can be illustrated via the use of a Punnett square:

F1 gТЗЕУ gТЗЕУ GT5Er GgT5ТЗЕГЕУ GgT5ТЗЕГЕУ GT5Er GgT5ТЗЕГЕУ GgT5ТЗЕГЕУ Genotype - GgT5ТЗЕГЕУ Phenotype - Grey banshees with 3 tendrils and orange eye color.

Q1D)

a)

genotype Gg T5 T3 E'EY gg T T EYEY X phenotype Grey banshee with 3 tendrils and orange eyes; heterozygous at all loci White banshee with 7 tendrils and yellow eyes; homozygous at all loci

b) Since the genes for color and tendril count are linked, they do not assort independently (this can be represented by overwriting them with a bar). Hence G and T5 would stat together and g and T3 would stay together in the gametes of F1. Since we are ignoring the eye color gene, we can construct a Punnett square to compute all the possible allele combinations resulting from this cross:

As you can see, none of the offspring will have white coat color and 5 tendrils as all individuals homozygous for g (white) carry the T3 and T7 alleles, and since T3 is dominant to T7, they all will have 3 tendrils. Those individuals with 5 tendrils on the other hand will all be grey. So, 0% of the offspring will be white with 5 tendrils.

F1 g17 gT7 GT5 GgT5T7 GgT5T7 gT3 ggT3T7 ggT3T7

c)

GT5Er GT5Ey GT3Er GT5Ey GT7Ey GgT5T7ErEy GgT5T7EyEy GgT3T7ErEy ggT3T7EyEy GT7Ey GgT5T7ErEy GgT5T7EyEy GgT3T7ErEy ggT3T7EyEy
Here I have constructed a Punnett square to better understand the result of the cross between the F1 banshees and a true breeding white banshee with seven tendrils and yellow eyes ($PLtc3YzH8X8RvLrwwfYAAAAABJRU5ErkJggg==$ EyEy). From this cross, you can see that 1/4^th (0.25) of all banshees from this cross will be grey with 3 tendrils and orange eyes.

I hope this helps :)