Question

You have the great fortune to have been given a miniature dragon. These dragons are prolific...

You have the great fortune to have been given a miniature dragon. These dragons are prolific and rapid breeders. Your dragon has magenta scales and green fire breath which are known to be inherited traits and are coded for by two different loci (genes) with two alleles each You don't know what the genotype of your new dragon is and you want to figure it out so you perform a testcross with your friends silver scaled, red fire breathing dragon.

From this test cross you obtain the following genotypes of baby dragons

Magenta scales, green fire - 44 dragons

silver scales, green fire - 59 dragons

magenta scales, red fire - 56 dragons

silver scales, red fire - 41 dragons

a) From the information given which alleles for each gene are dominant and       which are recessive?

b) What was the genotype of your friend’s dragon?

c) What are all of the possible genotypes of your dragon?

d) Test these possibilities against the observations from the testcross using a chi-squared test (Note: you do not have to do a chi-squared for all possible genotypes, but you do have to come up with the correct genotype).

e) What is the genotype of your dragon? Use the evidence from the results of your cross and your chi-squared test to support your claim.

Homework Answers

Answer #1

My dragon = magenta, green = M_G_

Other dragon = silver, red = mmgg

Ratio obtained after test cross -

44/41, 59/41, 56/41, 41/41

1.07, 1.43, 1.36, 1

The ratio is approximately 1:1:1:1. This is obtained when the dominant genotype is heterozygous at both the loci.

Total # progenies = 56 + 59 + 41 + 44 = 200

Expected # = 200/4 = 50

A. mmgg

B. MMGG, MmGG, MMGg, MmGg

C.

Null hypothesis - The genotype of my dragon is MmGg.

Phenotype #observed #expected (O-E)^2/E
Magenta green 44 50 0.72
Silver green 59 50 1.62
Magenta red 56 50 0.72
Silver red 41 50 1.62

Calculated chi square = 0.72 + 0.72 + 1.62 + 1.62 = 4.68

Degrees of freedom = 4 - 1 = 3

p value = 0.9 to 0.95

Null hypthesis accepted.

The genotype is MmGg.

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