GenCh4answers - 1. You perform the following cross and are...

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1. You perform the following cross and are told that the two genes are 10 m.u. apart. A B/a b × a b/a b Among their progeny, 10 percent should be recombinant ( A b/a b and a B/a b ) and 90 percent should be parental ( A B/a b and a b/a b ). Therefore, A B/a b should represent 1 / 2 of the parentals or 45 percent. 2. P A d / A d × a D / a D F 1 A d / a D F 2 1 A d / A d phenotype: A d 2 A d / a D phenotype: A D 1 a D / a D phenotype: a D 3. P R S/r s × R S/r s gametes 1 / 2 (1 – 0.35) R S 1 / 2 (1 – 0.35) r s 1 / 2 (0.35) R s 1 / 2 (0.35) r S F 1 genotypes 0.1056 R S/R S 0.1138 r s/r S 0.1056 r s/r s 0.1138 r s/R s 0.2113 R S/r s 0.0306 R s/R s 0.1138 R S/r S 0.0306 r S/r S 0.1138 R S/R s 0.0613 R s/r S F 1 phenotypes 0.6058 R S 0.1056 r s 0.1444 R s 0.1444 r S 5. Because only parental types are recovered, the two genes must be tightly linked and recombination must be very rare. Knowing how many progeny were looked at would give an indication of how close the genes are. 6. The problem states that a female that is A/a · B/b is test crossed. If the genes are unlinked, they should assort independently and the four progeny classes should be present in roughly equal proportions. This is clearly not the case. The A/a · B/b and a/a · b/b classes (the parentals) are much more common than the A/a · b/b and a/a · B/b classes (the recombinants). The two genes are on the same chromosome and are 10 map units apart. RF = 100% × (46 + 54)/1000 = 10% A B a b 10 m.u.
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56 Chapter Four 7. The cross is A/A · B/B × a/a · b/b . The F 1 would be A/a · B/b . a. If the genes are unlinked, all four progeny classes from the test cross (including a/a ; b/b ) would equal 25 percent. b. With completely linked genes, the F 1 would produce only A B and a b gametes. Thus, there would be a 50 percent chance of having a b/a b progeny from a test cross of this F 1 . c. If the two genes are linked and 10 map units apart, 10 percent of the test cross progeny should be recombinants. Since the F 1 is A B/a b , a b is one of the parental classes ( A B being the other) and should equal 1 / 2 of the total non-recombinants or 45 percent. d. 38 percent (see part c) 8. Meiosis is occurring in an organism that is C d/c D , ultimately producing haploid spores. The parental genotypes are C d and c D , in equal frequency. The recombinant types are C D and c d , in equal frequency. Eight map units means 8 percent recombinants. Thus, C D and c d will each be present at a frequency of 4 percent, and C d and c D will each be present at a frequency of (100% – 8%)/2 = 46%. a. 4 percent b. 4 percent c. 46 percent d. 8 percent 10. a. The three genes are linked. b. Comparing the parentals (most frequent) with the double crossovers (least frequent), the gene order is v p b . There were 2200 recombinants between v and p , and 1500 between p and b . The general formula for map units is m.u. = 100%(number of recombinants)/total number of progeny Therefore, the map units between v and p = 100%(2200)/10,000 = 22 m.u.,
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This note was uploaded on 09/25/2010 for the course BIOL 3301 taught by Professor Gunaratne during the Fall '05 term at University of Houston.

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GenCh4answers - 1. You perform the following cross and are...

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