Download Answers to Quiz 3:

Answers to Quiz 3:
1. Interval 31B5- 31C1 is the only region common to all deletions, defining the region
where gene x is found:
2. For these questions, the recessive mutations uncovered by the same deletion must be
located near each other. The logic is to start with the smallest mutations, then use others
to build the least ambiguous map:
(a,b)
(a,b,d)
abd or dab
cabd or dabc
(a,b,c)
cabdf
(b,d,f)
(d,e,f)
cabdfe
or
efdbac
Ans: (a)
3. Each individual chromosome will represent a separate linkage group, and since the
species are unrelated, the number should equal the haploid gene complement of both
genomes (22 + 9 = 31 linkage groups). Ans: (b).
4. The two species have four chromosomes that can pair- these are the four bivalents, but
two sets of homologs come together to form a tetravalent. This is most probably due to a
reciprocal translocation in one species, where the linkage groups found on two different
chromosomes have exchanged genes. Ans: (a)
5. This type of cancer is (usually) due to the inappropriate expression of the c-myc gene,
placed under the control of a transcriptional regulator (IgH enhancer) expressed in
lymphocytes, in the immunoglobulin gene cluster. See pp. 492-494 in text. Ans: all of the
above (d).
Fetus
Mrs.
Mr.
Simpson Simpson
6-8.
Fetus
Mr.
Mrs.
Simpson
Simpson
Fetus
Mrs.
Mr.
Simpson Simpson
The problem is with Mr. Simpson, who is heterozygous for a pericentric inversion. A
crossover within the inversion loop formed between the two chromosome six homologs
in meiosis one will generate a chromosome with duplications and deficiencies.
6. The chromosome was derived from the father, due to a crossover between homologs
within the inversion loop. Ans: (a)
7. The chromosome in the child contains duplications and deficiencies. Ans: (d)
8. False. It is possible that a gamete could be generated that did not contain a single
crossover within the inversion loop.
9. Oncogenesis can be caused by mutation. All these forms of mutation can be linked to
specific types of cancers…see text. Ans: (e).
10. This experiment describes the process of bacterial conjugation. This mode of bacterial
gene transfer requires cell-cell contact from a donor (F+) to a recipient (F-) cell. Ans: (a).