You have been characterizing the gene responsible for chronic hereditary ear hai
ID: 91956 • Letter: Y
Question
You have been characterizing the gene responsible for chronic hereditary ear hair disorder by mapping restriction enzyme sites in the gene. The coding region of the gene contains a single Hind III site, but no Eco RI sites (there will be sites for both outside the gene in the other genomic DNA). At this point, you want to clone the gene so you can study it further and use it as a probe. The plasmid you want to use contains the ampicillin resistance gene, which has a Hind III site within it, and the tetracycline resistance gene, which has an Eco RI site within it. Describe the steps you would take to clone the intact gene. Chronic hereditary ear hair disorder (CHEHD) is a recessive trait. The recessive allele differs from the wild-type allele by only a single point mutation, which conveniently falls within the Hind III site. The mutation changes the DNA sequence within the site such that Hind III does not recognize and cut the sequence. Utilizing this information, you obtain blood samples from five newborn subjects, and isolate their DNA. You cut the DNA with Hind III, and perform a Southern blot on the DNA, using your cloned gene as a probe. The following bands show up on the Southern blot: State the genotype of each of the newborns, and indicate which you expect to develop the heartbreak of CHEHD, and why.Explanation / Answer
First part:
From RNA, we have to make cDNA.
We should design primers that will selectively amplify coding region of the gene from cDNA. We should add EcoRI restriction site sequence to the 5’ end of both primers.
With such primers, we have to do PCR taking cDNA as template.
The PCR product should be run on agarose gel and band of appropriate size should be excised.
Gel extraction should be performed to isolate amplified product.
This should be digested with EcoRI and loaded on agarose gel. Digested product should be purified using gel extraction.
Digested product will have sticky ends on both sides.
Similarly, vector should be cut with EcoRI and purified from the gel.
Linearized vector and the purified PCR product should be ligated and transformed into bacteria.
Recombinants can be selected by plating the cells on ampicilin containing media. The colonies will then be inoculated on tetracycline containing media. Colonies did not survive on tetracycline media but grew in ampicilin media are recombinant clones.
Second part:
The wild type gene has HindIII restriction site. Mutation disrupts it and it will no longer be cut with HindIII.
Every individual has two copies of each gene. CHEHD is recessive trait vcaused by mutation. It means, individual with both copies of the gene mutated is going to develop CHEHD. The one with only one copy or no copy of mutated allele will be normal.
If the person has two copies of wild type gene, both copies will be cut by HindIII. This cleavage produces two fragments which will be seen on southern blot as two bands. Therefore, if we see two bands, the individual is homozygous for wildtype allele. They will not have CHEHD.
If the person has two copies of mutated gene, both copies will not be cut by HindIII as mutation disrupts HindIII site. Since there is no cleavage, they will show one bad on blot. So, if an individual shows single band, it suggests that the individual is homozygous for mutant allele. They will have CHEHD. Please note that this single band appears at higher molecular weight compared to the two bands seen in wildtype homozygous individuals.
If the person has one each copy of wild type and mutant alleles, there will be three bands. Bottom two bands (low molecular weight) are because of cleavage of wild type allele and the top band (high molecular weight) is because of mutant allele. They will not have CHEHD
As per the picture, individuals 1 and 3 have CHEHD (homozygous for mutant allele).
Individuals 2 and 5 will not have CHEHD (homozygous for wild type allele)
Individual 4 will also not have CHEHD (heterozygous)
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