Errors in Solution Manual Chapter 4 Chapter 7 - Harvard
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Errors in Solution Manual Chapter 4 Chapter 7 - Harvard
Errors in Solution Manual. (Corrections have been underlined.) Chapter 4. 7. The
cross is A/A · B/B χ a/a · b/b. The F1 would be A/a · B/b. The testcross would be ...
Errors in Solution Manual (Corrections have been underlined.)
Chapter 4 7.
The cross is A/A · B/B × a/a · b/b. The F1 would be A/a · B/b. The testcross would be A/a · B/b × 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 F1 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 F1 .
c.
If the two genes are linked and 10 map units apart, 90 percent of the test cross progeny should be parentals. Since the F1 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 parentals or 45 percent.
d.
38 percent (see part c)
Chapter 7 21.
a.
The top strand.
.
b.
The replication fork should be on the right side. (i.e. image should be flipped left to right, disregarding the sequence). .. .. 5’DNA polymerase AT TA TCG AG TA CA CG UG AT CU CG -5 A C ’ T ACTGACAGTC-3’.... G T RNA primer C A C TGACTGTCAG-5’.... CG TGA T C GA G AC CTA T G G TC CAT T A AG 5’-- T A . . ’ .. ..3 ..
c.
....5’-ATTCGTACGATCGACTGACTGACAGTC-3’.... ....3’-TAAGCATGCTAGCTGA CTGACTGTCAG-5’.... ....5’-ATTCGTACGATCGACT GACTGACAGTC-3’.... ....3’-TAAGCATGCTAGCTGA CTGACTGTCAG-5’.... d.
Yes, but the other replication fork would be moving in the opposite direction and the bottom strand, as drawn, would now be the leading strand and the top strand would now be the lagging strand.
Chapter 10 28.
The last line should read “cells are lacI–, lacIs, lacO–, and crp–.”
Chapter 11 7.
(a) clone the mutant by using convenient restriction sites flanking the actin gene and then sequence it. In order to do PCR, at least some of the sequence of the actin gene must be known. In the problem, only the position of the actin gene is known. If the position of the gene and the sequence of the gene is known, doing PCR is easier because it requires fewer steps and is much faster.
Chapter 15 10.
The likely origin of a disomic is nondisjunction during meiosis. Depending whether the nondisjunction took place during the second or first division, you would expect one nullosomic, or two nullosomics and another disomic, respectively.
