Problem Set 2
This problem set covers material from January 19
through January 26
From Griffiths et al.:
Chapter 2, problems 4, 8, 20, 22, 24, 28, 29, 30,
38, 39, 46, 47, 48,52, 53, 64
Chapter 6, problems 7, 8, 9, 12, 13, 14, 16, 23
2.4) Because the DNA levels vary four-fold, the range covers cells that are haploid (gametes) to
cells that are dividing (after DNA has replicated but prior to cell division).
The following cells
would fit the DNA measurements:
diploid cells in G1 or cells after meiosis I but prior to meiosis II
diploid cells after S but prior to cell division
2.8) In large part, this question is asking, why sex?
Parthenogenesis (the ability to reproduce
without fertilization – in essence, cloning) is not common among multicellular organisms.
Parthenogenesis occurs in some species of lizards and fish, and several kinds of insects but it is
the only means of reproduction for only a few of these species.
In plants, about 400 species can
reproduce asexually by a process called apomixis.
These plants produce seeds without
However, the majority of plants and animals reproduce sexually.
reproduction produces a wide variety of different offspring by forming new combinations of
traits inherited from both the father and the mother.
Despite the numerical advantages of asexual
reproduction, most multicellular species that have adopted it as their only method of reproducing
have become extinct.
However, there is no agreed upon explanation of why the loss of sexual
reproduction usually leads to early extinction or conversely, why sexual reproduction is
associated with evolutionary success.
On the other hand, the immediate effects of such a
scenario are obvious.
All offspring will be genetically identical to their mothers, and males
would be extinct within one generation.
2.20) Do a testcross (cross to a/a).
If the fly was A/A, all the progeny will be phenotypically A;
if the fly was A/a, half the progeny will be A and half will be a.
2.24 a) You do not expect the mutation to be recessive.
This would be an example of a
haploinsufficient gene since one copy of the wild-type allele does produce enough protein
product for normal function. b) An important assumption would be that having five of eight units
of protein product would result in an observable phenotype.
It also assumes that the regulation
of the single wild-type allele is not affected.
Finally, if the mutant allele was leaky rather than
null, there might be sufficient protein function when heterozygous with a wild-type allele.