One recessive mutation in human chromosome 12 can cause a rare disease that lead
ID: 57847 • 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 20 cM from the cetromere. A RFLP marker is identified to be as close as 5cM 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: https://media.cheggcdn.com/media/d2d/d2d6ad82-3767-4b98-924e-403ab3d6c30d/phpfXBgIt.png Recombination needs to be taken into consideration for the answers. 1. Before Mary does genetic test for this RFLP marker, what is the probability of Mary as a carrier of the death gene and being a heterozygote for the RFLP marker? 2. After Mary did her RFLP test and found out she was indeed a heterozygote for the marker, then what is the probability of Mary 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 10cM 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 the SNPs (that is to say, one chromosome is wild-type and another one contains the SNP). 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? 4. What is the probability of Sister 3 as a carrier of the death gene?
Explanation / Answer
Before the RFLP, it can be concluded that both the parents of Mary are heterozygous and the probability that Mary is a carrier is 2/3. Since Mary is not affected, she cannot be homozygous recessive. Therefore, she must be either homozygous dominant or heterozygous. Hence, the probability is 2/3
After the RFLP, it is clear that from the restriction digestion that Mary is heterozygous carrier. Hence, the probability after RFLP analysis is 1.
Since the gene is 9 cM away, the probability of recombination is 9+9 = 18% (from both the parents). Therefore, the probability that Mary is a carrier of the death gene is 0.18
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.