1) A)Using the genotype provided above, calculate the number of unique gametes p
ID: 207937 • Letter: 1
Question
1)
A)Using the genotype provided above, calculate the number of unique gametes produced by this horse.
B)How many will transmit all three disease alleles present in this horse?
C)How many gametes will carry just 1 of the 3? How many gametes will not transmit any disease alleles?
Dam Dam Name: GS Cookie Breed: Registration Phenotype: Genetic Disorders XHYPP N/N No HYPP alleles X HERDA N/HRD HERDA carrier X GBED N/N No GBED alleles X PSSM1 N/N No PSSM1 alleles X MH N/N No MH alleles XJEB N/N No JEB alleles XCA N/CA CA carrier X LFS N/LFS LFS carrierExplanation / Answer
A) According to the question no. of heterozygous gene pairs= 3 (N/HRD, N/CA & N/LFS)
Now, according to the formula if no. of heterozygous gene pairs is n, then no. of unique gametes formed is 2n....(1)
Putting values in eqn. (1), we get-
No. of unique gametes = 2n = 23 = 8
B & C) We know that F2 trihybrid cross resulting in 27:9:9:9:3:3:3:1 phenotypic ratio.
So, for N/HRD, N/CA & N/LFS we will get-
NNN= 27/64
NN/HRD= 9/64
NN/CA= 9/64
NN/LFS= 9/64
N/CA/HRD= 3/64
N/CA/LFS= 3/64
N/LFS/HRD= 3/64
CA/HRD/LFS= 1/64
So, probability of all 3 disease alleles transmission (CA/HRD/LFS) = 1/64 (In other words 1 in every 64 gametes will transmit all 3 disease alleles).
So, probability of only 1 allele transmission (NN/HRD, NN/CA, NN/LFS) = 9/64 + 9/64 + 9/64 = 27/64 (In other words 27 in every 64 gametes will transmit only 1 disease allele).
So, probability of not transmit any disease (NNN)= 27/64 (In other words 27 in every 64 gametes will transmit any disease alleles).
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