4. At time 0 an Hfr strain was mixed with an F- strain, and at various times aft
ID: 300853 • Letter: 4
Question
4. At time 0 an Hfr strain was mixed with an F- strain, and at various times after mixing, samples were removed and agitated to separate conjugating cells. The cells were then plated onto medium containing streptomycin and the number of prototrophic recombinants for each of the markers was determined. The cross may be written as Hfr: a b'c'd'efg'h'str F ab d ef gh str (NO ORDER IS IMPLIED The graph of the number of recombinants against time is shown below 100 ht %str, r ecombinants 7S so 25 S 10 15 20 25 Minutes after star t of conj ugati on Based on these data, which of the following statements is TRUE A. h' and e are farther apart on the bacterial chromosome than b' and a B. Most e strr recombinants are likely to be Hfr cells C. All F cells which received a from the Hfr in the chromosome transfer process must also D. The order of gene transfer from Hfr to F is FIRST a' then g', then b', then e, then b' have received b. 5. In a population at Hardy-Weinberg equilibrium for two alleles B and b at one locus, the frequency of the recessive allele b is 0.2. What % of individuals display the dominant phenotype for this gene/trait? A. 4% B. 32% C. 64% D. 80% E. 96%Explanation / Answer
100 % recombination has occured when the curve is flat. For exapmle, h+ has transferred around 8th minute. So the sequence is : h+ e+ b+ g+ a+.
Analyzing each given statement:
4. A. From the above sequence, it is clear that e+ and h+ are located closer that b+ and a+. This statement is incorrect.
B. From the graph, we can find that e+ is fully transferred around 14 minutes and a+ is fully transferred around 25 minutes. So the cells would not be 100% Hfr even if cells are 100% e+. This is also incorrect.
C. From the sequence, we can interpret that a+ is transferred after the transfer of b+. So all the a+ cells must be b+. This is correct.
D. The sequnce of trnafer is given ablove. The sequence given in the problem statement is incorrect.
ANSWER: C
5. Frequency of recessive allele (q) = 0.2
Frequency of dominant allele (p) = 1 - q = 1- 0.2 = 0.8
According to Hardy-Weinberg equilibrium, p2+2pq+q2 = 1
Frequency of individuals displaying dominant phenotype =p2+2pq =0.82+(2*0.8*0.2) =0.64+0.32 = 0.96
Frequncy of individuals displaying dominant phenotype = 0.96 = 96%
ANSWER: E.
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