A00%20final%20exam%20vs2%20KEY

# A00%20final%20exam%20vs2%20KEY - BICD100 Final Exam(A00 VS...

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BICD100 Final Exam 3/19/08 (A00): VS. 2 KEY (PAGE NUMBERS LOWER RIGHT CORNER) 1.) In the 16 th century, a Norwegian Viking ship crashed into the cliffs along the coast of Greenland during a storm. The fifty survivors who made it to shore took up residence in nearby villages. Prior to this incident, the population of Norway had been ravaged for several generations by the bubonic plague. Because of this, the Norwegian population from which the Vikings came had a relatively high frequency of individuals with genetic resistance to plague infection due to a dominant allele at a single gene (allele frequency 0.5). The 300,000 people living in this part of Greenland when the Vikings arrived, who had never been exposed to the plague, had a very low frequency of resistance alleles (10 -5 ) arising from spontaneous mutations. Let A represent the resistance allele (dominant) and a represent the sensitive allele (recessive). f(A) = p and f(a) = q a. Assuming random mating between the Vikings and Greenlanders, what was the frequency of plague resistant individuals in the next generation (10 pts)? m = 50/300,050 = 0.000167 p i = 1 x 10 -5 p m = 0.5 p' = mp m + (1 – m)p i = (0.000167)(0.5) + (1 – 0.000167)(1x10 -5 ) = 0.0000935 or 9.35x10 -5 q’ = 1-p’ = 0.9999065 plague resistant individuals are AA or Aa f(AA) + f(Aa) = p’ 2 + 2p’q’ = 8.7x10 -9 + 0.000187 = 0.000187 or 1.87x10 -4 b. Assuming continued random mating, no further migration in or out of the region, no new resistance alleles arising through mutation, no selection for or against resistance alleles, and no genetic drift, what was the frequency of plague resistant individuals 4 generations (100 years) later (4 pts)? This fulfills all the requirements for Hardy-Weinberg equilibrium so allele frequencies and genotypic frequencies will remain the same as in a. c. 100 years later, arrival of another ship from Europe carrying plague-infected rats resulted in introduction of the plague into this part of Greenland. If genetically susceptible individuals on average had only half as many progeny as resistant individuals due to a high rate of plague deaths, what was the frequency of plague resistant individuals in the following generation (10 pts)? W for aa = 0.5 so s = 0.5 f(a) = q is the frequency of sensitive allele, which is the recessive one that is selected against here. q' = q – sq 2 /1 – sq 2 = 0.9999 – (0.5)(0.9999) 2 /1 – (0.5)(0.9999) 2 = 0.9998 p’ = 1 – q’ = 0.0002 or 2x10 -4 New f(AA) + f(Aa) = p’ 2 + 2p’q’ = 4 x 10 -8 + 0.0004 = 0.0004 or 4x10 -4 1

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2.) A Neurospora strain of mating type A carrying the arg1 mutation (requires arginine in the medium to grow) was crossed to another strain of mating type a that can grow without arginine (+). 400 octads were analyzed, yielding the results shown below (the order of spores in the octad is shown from top to bottom). Draw a map illustrating the linkage relationships of mating type and arg1 genes to the centromere(s) and to each other, putting numbers on each distance in cM (20 pts).
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