8. Here\'s a less fun scenario. A couple is expecting twins and has ju heterozyg

Question

8. Here's a less fun scenario. A couple is expecting twins and has ju heterozygous carriers for two autosomal recessive traits, albinism and sickle cell anemia. st learned that they are both f they have fraternal twins, what is the probability that both twins will be phenotypicaly normal and kte one will be a carrier for only the albinism allele and the other will be a carrier for only thes ypi ickle cell allele? (Hint: to solve this problem you need to use both the sum and product laws but do not need to mess with binomials. Also, notice that which twin has which genotype isn specified.) (2pts)

Explanation / Answer

Let the Genotypes of the carrier parents be Aa for albinism and Ss for sickle cell anaemia

so, parent 1 has a genotype AaSs and parent 2 has genotype AaSs. Since,both the parents are heterozygous for the diseases.

The events must take place: the twins must receive Aa alleles, and it must receive Ss alleles. For autosomal recessive traits, the offspring must receive one copy of the affected gene from each carrier parent; it should be homozygous for the trait.

Considering,the two events to be independent because the genes assort independently and do not affect one another .

Using Punnet Square for each disease,

For Albinism,

AA-normal; Aa- carrier ;aa- affected

For Sickle cell Anaemia.

Ss-normal; Ss- carrier ;ss- affected

In order to find the probability of twins for normal phenotype, we have to consider the fact that they are genotypically carriers of the two given traits, and none of the homozygous recessive forms are present in their genotype.

For albinism such genotype is Aa and AA, where the individual is a carrier an/or not affected, Phenotypically normal.

Probability of getting Aa =2/4= 1/2   Probability of getting AA = 1/4

For sickle cell anaemia such genotype is Ss and SS, where the individual is a carrier but not affected.

Probability of getting Ss =2/4= 1/2 Probability of getting SS = 1/4

Therefore probability of getting AaSs = 1/2 x1/2 = 1/4

probability of getting AASS = 1/4 x1/4 = 1/16 (unaffected offsprings)

Next , probability for only carriers of albinism.

the genotype should be: AaSS,

Probability of getting Aa =2/4= 1/2   Probability of getting SS =1/4= 1/4

probability of getting AaSS = 1/2x1/4 = 1/8

probability for only carriers of sickle cell anaemia.

the genotype should be AASs.

probability for SS = 1/4    probability for Ss = 2/4=1/2

Therefore  probability for only carriers of sickle cell anaemia = 1/4x1/2 =1/8.

A a A AA Aa a Aa aa

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