One recessive mutation in human chromosome 12 can cause a rare disease that lead
ID: 55919 • Letter: O
Question
One recessive mutation in human chromosome 12 can cause a rare disease that leads to death within the first year. This gene is located approximately 30 cM from the cetromere. A RFLP marker is identified to be as close as 3cM to this lethal gene. The RFLP markers results from the presence or absence of a BamHI restriction enzyme site. The presence of the BamHI site results in the production of 5kb and 3 kb restriction fragment that can be visualized on an agarose gel. The absence of the HindIII site results in an 8.0Kb fragment. Mary came from a family, which unfortunately carry this horrible gene. One of Mary sisters passed away in her 6-month age. Mary is now 25 and planning to have a child. She is concerned about passing this recessive allele to her child so she collected DNA from everyone in her family to do the RFLP test. The results are shown below:
1. What is the probability of Mary as a carrier of the death gene?
2. What is the probability of Sister 3 as a carrier of the death gene?
3. Mary wants to further confirm if she is a carrier for this lethal gene. She was told that a new SNP marker was identified at a location of 4cM from the lethal gene and 7cM from the previously identified RFLP marker. She requested to do SNP test for all the DNA samples. The results are: her parents are heterozygous for SNPs. The sister who passed away is homozygous for the SNP. Mary does not have this SNP. So now what is the probability of Mary as a carrier of the death gene?
Fa ther Mother Sister1 Sist er2 Sister3 Mary RFLPS 8kb 5kb 3kb Sister1 Sister2 Sister3 MaryExplanation / Answer
For RFLP marker 1: Let us denote the gene allele for recessive allele as “a” and dominant allele as “A”
Now from the RFLP results, both the parents show heterozygosity for the marker 1, which means they have the genotype “Aa” each. From the RFLP results, it is also evident that Mary is heterozygous for the marker, which means Mary also carries the genotype Aa. We can also be sure that the Sister 3 does not carry the death marker as she shows both the alleles for the wild type “AA” (no 5+3 band seen on RFLP).
To calculate probability, we will draw the Punnett Square:
Parents genotype
A
A
A
AA
Aa
a
Aa
Aa
1) Here, we can see that the Aa genotype is possible only 2/4 times, i.e ½ the times. Hence the probability of Mary carrying the death gene is ½
2) As the Sister 3 does not carry the allele for the death gene, the probability is 0.
3) Now if we include a second marker ( recessive allele b and dominant allele B), we get the following genotypes for the parents, and Mary.
Parents: AaBb, Mary: AaBB
The Punnett square results are as follows:
Parent’s genotype
AB
Ab
aB
ab
AB
AABB
AABb
AaBB
AaBb
Ab
AABb
AAbb
AaBb
Aabb
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
The probability of Mary carrying the deceased gene is2/16 or 1/8 of the times.
Parents genotype
A
A
A
AA
Aa
a
Aa
Aa
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