A mother is heterozygous for an over-expressed plastin allele on her X chromosom

Question

A mother is heterozygous for an over-expressed plastin allele on her X chromosome, thus has longer than normal axons. She is also heterozygous for the SMN1 gene on chromosome 5. Her children’s father is normal for plastin and has normal length axons. He too is heterozygous for the SMN1 gene. Both genes influence spinal muscular atrophy type 1 (SMA1); two abnormal copies of the SMN1 gene are required to cause spinal muscular atrophy, but their effect is prevented by high expression of the plastin gene (Lewis 92).

A. Give the genotypes of the mother and the father; provide a legend for your symbols. For each gene, explain why the abnormal alleles are dominant or recessive.       (4 marks)

B. List all the possible genotypes and proportions of their offspring for the plastin gene.                                               (1 mark)

C. List all the possible genotypes and proportions of their offspring for the SMN1 gene.                                               (1 mark)

D. Showing your work, determine the probability that this couple will have a daughter with SMA.                                           (2 marks)

E. Showing your work, determine the probability that this couple will have a son with SMA.      

PLEASE DO NOT ATTEMPT UNLESS YOU ARE A GENETICS EXPERT!!!

Explanation / Answer

Hi,

The mother is heterozygous for the plastin gene on X chromosome. So her genotype is XpX.

She is also heterozygous to SMN1 on chromosome 5. So her genotype is = Ss XpX

The father is normal for X chromosome and heterozygous forSMN1. = SsXY

Two recessive copies ‘ss’ can cause muscular atrophy. But presence of Xp can suppress it.

Genotype of father = SsXY

The abnormal gene for SMN1 is recessive because two copies are required to cause muscular atrophy. The abnormal gene of plastin is dominant because it can suppress the other SMN1 gene and only one copy suffices.

When they mate the possible offsprings are:

Mother = Ss XpX: Gametes = SX, SXp , sXp , sX.

Father = SsXY : Gametes = SX, SY, sX, sY

SX

SXp

sXp

sX

SX

SS XX

Normal female

SS XXp

Normal, Plastin , female

Ss XXp

Normal,plastin, female

SsXX

Normal, nornmal female

SY

SS XY

Normal normal male

SS Xp Y

Normal plastin male

Ss Xp Y

Normal, plastin male

Ss XY

Normal normal male

sX

Ss XX

Normal normal female

Ss X Xp

Normal plastin female

ss X Xp

Normal plastin female

ssXX

SMA, normal female

sY

Ss XY

Normal normal male

Ss XpY

Normal plastin male

ss XpY

normal, plastin,male

ss XY

SMA normal male

Answers for B and C are solved in the above table.

D. determine the probability that this couple will have a daughter with SMA.

From the table, out of 16 possibilities, the one with ssXX has SMA and is a female

So probability = 1 / 16.

E. determine the probability that this couple will have a son with SMA.      

From the table, out of 16 possibilities, the one with ssXY has SMA and is a male.

So probability = 1 / 16.

     

SX

SXp

sXp

sX

SX

SS XX

Normal female

SS XXp

Normal, Plastin , female

Ss XXp

Normal,plastin, female

SsXX

Normal, nornmal female

SY

SS XY

Normal normal male

SS Xp Y

Normal plastin male

Ss Xp Y

Normal, plastin male

Ss XY

Normal normal male

sX

Ss XX

Normal normal female

Ss X Xp

Normal plastin female

ss X Xp

Normal plastin female

ssXX

SMA, normal female

sY

Ss XY

Normal normal male

Ss XpY

Normal plastin male

ss XpY

normal, plastin,male

ss XY

SMA normal male

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