5CPractice_Final_answers - 1 The half-life of Carbon-14...

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1. The half-life of Carbon-14 decaying to Nitrogen-14 is 5,730 years. A mammoth bone fossil found in Siberia appears to be 22,920 years old. If these fossils now contain 6 grams of C-14, how much C-14 did they contain when this mammoth died? 22,920 6 17,190 12 11,460 24 5,730 48 1 96 22,920 / 5,730 = 4 half lives OR end = starting amount / 2 ^ half-lives 6 = start / 2 ^ 4 6 = start / 16 96 = starting amount 2. A disease is controlled by a single gene with two alleles. Assume Hardy-Weinberg equilibrium for this gene. If one in 2,000 monkeys have albinism, what is the ratio of heterozygous carriers to those with the disease? Heterozygous = Cc = 2pq Disease = cc = q^2 q 2 = 1/2000 = 0.0005 q = sqrt(0.0005) = 0.0224 p + q = 1 p = 1 - q = 0.976 Carrier = 2pq = 0.976*0.0224*2 = 0.0437 0.0437 / 0.0005 = 87.4 87 : 1 3. If a trait varies continuously then it's inheritance is most likely an example of __________ ? Since it varies continuously NOT discretely then it is an example of polygenic inheritance. Saying it varies continuously is another way to say it is a quantitative trait. Can you think of examples of quantitative traits? Discretely inherited traits (i.e. purple vs white flower color) are typically controlled by a single gene. 4. Jenny's sister has PKU, a recessive disorder. Neither of her parents have the disorder. What is the chance that Jenny is a carrier of the PKU allele? (Jenny does not have the disease) There is a 2/3 chance that Jenny is a carrier. Let's assume P is the wild-type or NON-Disease allele and p is the disease allele. Her parents must both be carriers (Pp) if they do not show symptoms and the sister (pp) has the disease. So if Jenny does not have the disease she can either be Heterozygous (Pp) like her parents or homozygous for the non-disease allele (PP). There are 2 ways to be Heterozygous (Pp) if the parents are both heterozygous (Do the Punnett square of Pp x
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Pp if this doesn't make sense to you) and 1 way to be homozygous dominant (PP). So there is a 2 out of 3 chance that Jenny is a carrier. 5. What is the chance that Jenny (from the previous question!) and her husband Steve who doesn't have the disease will have a child with PKU if the disorder is seen in 1/ 15,000 births? 0.00269 (or 0.27%) Chance that they will have a pp child is 1/4 if both the parents are Pp. What is the chance that Steve will have the disorder is the same as asking, what is the frequency of carriers in the population if I give you the frequency of the disease allele. Use Hardy-Weinberg equation here. p 2 + 2pq + q 2 = 1 and p + q = 1 Disease is seen 1 / 15,000 births. This means q 2 = 1/15,000 or so q = p + q = 1 so p=1 - q or p = 0.992 Carrier frequency is 2pq 2pq = 2 * 0.992 * 0.00816 = 0.0162 (1.62 % of population is a carrier).
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5CPractice_Final_answers - 1 The half-life of Carbon-14...

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