1. The family in the pedigree below contains many individuals with Huntington\'s

Question

1. The family in the pedigree below contains many individuals with Huntington's disease Huntinton's is a monogenic trait conferred by a dominant allele (H) of the huntingtin gene Wild type alleles (h) are recessive to the disease causing allele. Affected individuals are shaded in grev. Also shown are each individual's genotvpe for a single nucleotide polymorphism, SNP-1. There are two alleles of SNP-1 in this family, which differ by a single nucleotide, A or G. In answering the questions below, use H and h to indicate the mutant and non-mutant Huntington alleles, and SNP1-A and SNP1-G for the SNP alleles. Modified from Hartl and Ruvolo, 2011. A/G A/G A A/G A A/G A/G A/ A A/GA A A A/G A/G A A/G A A/G A A A/G A/G A) Assuming that the Huntington gene and the SNP-1 are unlinked, what genotypes (and in what ratios) would you expect among the offspring of I-1 x 1-2? B) You suspect that the Huntington gene and SNP-1 are linked. Use the data in the pedigree to estimate the recombination frequency between these two loci. Include recombinants in Generations II and III in your calculations. C) Calculate a lod score for the recombination frequency you estimated in part B

Explanation / Answer

We have chosen SNP1A for A type progeny and SNP1G for G type progeny. The disease is dominant type therefore, if any progeny is not infected will get hh genotype. Here no progeny will get HH genotype because from the 1st generation the progenies are uninfected also. If progenies are uninfected, it means one of their parents is homozygous recessive (for uninfected) and the second one wills the heterozygous.

Part A-

From the cross of I-1 and I-2 we will get-

SNP1A progenies = 4

SNP1G progenies =2

Infected =3

Uninfected = 3

Therefore, ratios are-

SNP1A : SNP1G =2:1

Infected: Uninfected =1:1

Part B-

If those genes are linked, then we have one progenies from I-1 and I-2 crosses who are having SNP1A and infection trait.

Recombination Frequency = Numbers of recombinant progenies / Total progenies

Recombination Frequency = 1/ 6 = 0.17

Part C-

LOD score = log [Probability of linkage birth sequence / Probability of Non-linkage birth sequence]

Here the Probability of linkage birth sequence means the probability of having a linked progeny. In the cross, we have five progenies who are parental type. Therefore, the probability of a single genotype will be=

(1-0.17)/5 = 0.17

The Probability of linkage birth sequence is the product of all parental probability and recombinant probability.

Therefore,

Probability of linkage birth sequence = (0.17)5 x 0.17 [here five parental types have 0.17 probability and the recombinant has the 0.17 probability.

Probability of linkage birth sequence = 0.00002413

Now

The probability of non-linkage birth types =

Because in case of unlinked genotype the recombination frequency of a gene is 0.5, therefore they genotype frequency would be 0.25.

Now The probability of non-linkage birth types = (0.25)6 = 0.00024414

Now the lod score = log[0.00002413 /0.00024414]

Lod score = log 0.988

From the lod score table log values are near to 1.088 log score. It reveals the recombination frequency at 0.1.

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