Answers to Chapter 2
17. The broad bean is diploid 2n=18. Each haploid chromosome contains 4 m of DNA. The average size
of each chromosome during metaphase of mitosis is 13 μm. What is the average packing ratio of DNA at
metaphase? How is the packing achieved?
Answer 17) Nine chromosome in a haploid set (2n=18), and total of 4 m of DNA therefore, 4/9 m long on
an average. At metaphase, their average length is 13μm, so average packing ration is 13* 10-6 m : 4.4 *
10-1 m.
18. If we call the amount of DNA per genome x name a situation in diploid organisms in which the
amount of DNA per cell is
a. x b. 2x c. 4 x
Answer 18) x= haploid cells, 2x= diploid cells in G1Phase or cells after meiosis I but prior to meiosis II
4 x= diploid cells after S but prior to cell division.
20. Name two key functions of meiosis.
Answer 20) The two key functions of meiosis are to halve the DNA content and to reshuffle the genetic
content of the organism to generate genetic diversity among the progeny.
27. Do you agree with Francis Galton’s principle that half of our genetic makeup is derived from each
parent, one quarter from our grandparents and one -eighth from each of our great grandparents, and so
forth.
Answer 27) Yes, half of our genetic makeup is derived from each parent; each parent’s genetic makeup is
derived half from each of their parents.
28. If children obtain half their genes from one parent and half from the other parent, why aren’t the
siblings’s identical?
Answer 28) Because the half inherited is very random, the chances of receiving exactly the same half is
extremely small.
34. If you had a fruit fly, that was of phenotype A, what test would you make to determine if the fly‘s
genotype was A/A or A/a.
Answer 34) Test cross will be performed (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 the progeny will be a.
36. Two black guinea pigs were mated for several years and produces 20 black and 9 white offspring.
Explain these results, giving genotypes of parents and progenies.
Answer 36) The progeny ratio is approximately 3:1, including classic heterozygous-by-heterozygous
mating. Since black (B) is dominant to white (b): Progeny will be 3 ( black) : 1 ( white)=1BB, 2 B/b and 1
bb.
41.A large screen of Aabidopsis plants turns up two mutant plants (A and B) that have no trichomes. Each
plant is crossed with the wild type plants in both cases next generation (F1) had normal trichomes. When
F1 plants were selfed, the resulting F2 were as follows-F2 from mutant A plant 602 normal trichomes and
198 no trichomes
From F2 of B mutant- plant 267 normal and 93 no trichomes.
a. What do these results show?
b. Is it possible to predict F1 from crossing the original mutant A with the original mutant B?
Answer 41) a) A and B mutant plants are homozygous for recessive alleles. Both F2 crosses give 3:1
ratios of normal to mutant progeny. b) No, we do not know whether A and B mutations are in the same or
different genes.
44. The ability to taste the chemical phenylthiocarbamide (PTC) is an autosomal dominant phenotype and
the inability to taste is a recessive phenotype. If a taster woman with a non taster father marries a taster
man who in a previous marriage had a non taster daughter, what is the probability that their first child will
be
1. A non taster girl
2.A taster girl
3. A taster boy.
What is the probability that their first two children will be tasters of either sex?
Answer 44) The probability of a taster is ¾ and a non taster is ¼.
a) 1) Now, p (non taster girl)* (p girl) =1/4*1/2=1/8.
2) p (taster girl)=p(taster)* p (girl) =3/4*1/2=3/8
3) p (taster boy) =p (taster)* p ( boy) =3/4*1/2=3/8
b) p (taster for first two children)=p ( taster for first child) * p (taster for second child) =3/4*3/4=9/16
47. Suppose that a husband and wife are both heterozygous for a recessive allele for albinism. If they have
dizygotic twins, what is the probability that both the twins will have the same phenotype for
pigmentation?
Answer 47) Because the parents are heterozygous, both are A/a. Both twins could be albino or both twins
could be normal. The probability of normal is 3/4 and albino is ¼.
p (both normal) + p (both albino)
p (first normal)* p (second normal) + p (first albino) * p (second albino)
=(3/4)*(3/4) + (1/4)*(1/4)=9/16+1/16=5/8.