Tai 發問時間: 科學動物學 · 8 年前

4 question about human genetic

1. a particular disease is suspected of being a genetic disease. a person with the disease tests positive for 15 DNA markers, each detected with a specific DNA probe, and negative for 20 other markers. what would be the next step in trying to determine which of the markers might be located near a gene responsible for the disease?

2. how many copies of a DNA fragment are synthesized in 20 rounds of amplification by PCR (assuming that each step works correctly) ? how many will be synthesized in 30 rounds?

3. a point mutation that substitutes a single base for another changes that codon but not the succeeding ones. if two bases that are next to each other are replaced by two different bases, how many codons will be altered? on what might your answer depend?

4. what is the risk of a child's having a recessive genetic trait when both parents are from a population in which the frequency of the recessive allele is 1 in 1000?

1 個解答

  • C C
    Lv 6
    8 年前

    Some of these questions are a bit vague... but here it is:

    1. The next step in trying to narrow down which markers are linked with the disease causing gene would be to check his relatives who do not have the disease or check unrelated individuals who do have the disease. Relatives who do not have the disease could share some of the 15 markers that the patient carries, so you can rule them out. The unrelated affected individuals could carry a subset of the 15 markers, so you can eliminate the markers not shared.

    (this is obviously a GROSS simplification... markers may span a huge region, the causal locus may have incomplete penetrance, etc. etc.)

    2. The number of DNA molecule doubles each round of PCR. So if you started out with 1 molecule of DNA, then after 20 rounds you would have 1 x 2^20 molecules. After 30 rounds you would have 1x2^30.

    3. Your answer depends on where in the codon these two substitutions lies. If the two substitutions are in the same codon, then at most one codon is altered. If the two substitutions are in two different codons, then they must be in the 3rd position of one codon and the 1st position of the second codon. Then it depends on the degeneracy of the codons to determine how many amino acid changes occur.

    4. If the recessive allele frequency is 1/1000, then by Hardy-Weinberg the probability that one parent is a carrier is 2*(999/1000)*(1/1000) ~ 0.002. Then the probability that BOTH parents are carrier is 0.002 * 0.002 = 4x10^-6. Finally, if both parents are carriers, there's a 1/4 chance the child will be affected, or 4x10^-6 * 1/4 = 1x10^-6.

    參考資料: I study human genetics (though... these questions are from such a long time ago)
    • 登入以對解答發表意見