Page 58 - Genetics_From_Genes_to_Genomes_6th_FULL_Part2

Problems 217

conversion did not occur during formation of the Section 6.6
octad? Explain. 35. Figure 6.31 shows four potential outcomes of site-
specific recombination that depend on the relative ar-
a + rangement of the target sites for the recombinase
enzyme. Could homologous recombination (in the ab-
a +
sence of specific target sites and recombinase) also
a + cause all of the same kinds of outcomes? If so, what is
so different about how geneticists would use homolo-
a gous recombination and site-specific recombination?

a + 36. Each of the substrates for site-specific recombination
listed (a–f) is also the product of site-specific recombina-
a
tion that occurred at a different one of these substrates.
a Match each substrate (a–f) with its product (a–f).
a. a linear DNA with two target sites in the same ori-
a
entation
b. a linear DNA with two target sites in opposite ori-
−
−
−
+
33. From a cross between e f g and e f g strains of entations
+
+
Neurospora, recombination between these linked c. a circular DNA with two target sites in the same
genes resulted in a few octads containing the following orientation
ordered set of spores: d. a circular DNA with two target sites in opposite
orientations
+
+
e f g + e. a circular DNA with one target site and a linear
+
+
e f g + DNA with one target site
−
+
e f g +
−
+
e f g + f. two circular DNAs each with one target site
−
−
e f g − 37. Problem 52 in Chapter 5 discussed the use of mitotic
−
−
e f g − recombination to study the function of a gene called
−
−
e f g − smc during mouse development. The idea was that
−
+
−
−
e f g − mitotic recombination in a cell of an smc /smc het-
erozygote could produce a clone of tissue derived
a. Where was recombination initiated? from a daughter cell that was homozygous for the
−
b. Did crossing-over occur between genes e and g? smc mutation. You could recognize the cells in this
Explain. clone by the absence of green fluorescence from GFP.
−
−
+
+
c. Why do you end up with 2 f : 6 f but 4 e : 4 e a. Mitotic recombination is a rare event, making it
−
and 4g : 4g ? difficult for researchers to find smc /smc clones
+
−
−
d. Could you characterize these unusual octads as MI to study. Explain why scientists might want to sub-
or MII for any of the three genes involved? ject the same mice to X-rays to increase the fre-
Explain. quency at which the desired clones could be found.
34. In Step 6 of Fig. 6.27, the resolvase enzyme almost At what stage of mouse development would the re-
always cuts all four strands of DNA in the double searchers expose the animals to X-rays?
Holliday junction intermediate: both blue strands and b. An even more efficient way to induce mitotic re-
both red strands. Another way of stating this fact is combination is to construct mice that include the
that the enzyme cuts the DNA at Holliday junctions Flp/FRT system. (Flp is a recombinase enzyme in
1 and 2 in different ways, represented by the yellow yeast cells that promotes site-specific recombina-
arrows at junction 1 and the green arrows at junction 2 tion between two identical copies of a particular 34
in the figure. But rarely, the resolvase enzyme instead base pair long DNA sequence called an FRT site.)
cuts the DNA at both Holliday junctions in the same Assuming that these mice have a transgene that
way (yellow arrows at both junctions or green arrows can specify the Flp recombinase protein, where
at both junctions). In other words, at both junctions, would you put the FRT sites relative to the smc
the same red strand and the same blue strand are cut. gene and to the GFP gene?
What would be the outcome of this rare resolvase en- c. In part (b), what (in general terms) would you have
zyme behavior? to do with the Flp-encoding gene to make sure that