3
The Chromosomal Basis of Inheritance
BASIC PROBLEMS
1.
The key function of mitosis is to generate two daughter cells genetically
identical to the original parent cell.
2.
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.
3.
Its pretty hard to beat several billions of years of evolution but it might be
simpler if DNA did not replicate prior to meiosis. The same events responsible
for halving the DNA and producing genetic diversity could be achieved in a
single cell division if homologous chromosomes paired, recombined, randomly
aligned during metaphase, and separated during anaphase, etc. However, you
would lose the chance to check and repair DNA that replication allows.
4.
In large part, this question is asking, why sex? Parthogenesis (the ability to
reproduce without fertilization — in essence, cloning) is not common among
multicellular organisms. Parthenogenesis occurs in a some species of lizards and
fishes, and several kinds of insects but it is tbe only means of reproduction in
only a few of these species. In plants, about 400 species can reproduce asexually
by a process called apomixis. These plants produce seeds without fertilization.
However, the majority of plants and animals reproduce sexually. Sexual
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.
42
Chapter Three
5.
As cells divide mitotically, each chromosome consists of identical sister
chromatids that are separated to form genetically identical daughter cells.
Although the second division of meiosis appears to be a similar process, the
“sister” chromatids are likely to be different. Recombination during earlier
meiotic stages has swapped regions of DNA between sister and nonsister
chromosomes such that the two daughter cells of this division typically are not
genetically identical.
6.
a.
Because the sister-chromatids have not separated, the cell is undergoing
meiosis I. As the dyads are at the poles, the cell is at the end of anaphase I.
b.
Each chromosome will have been replicated and each would contain two
sister-chromatids. The replicated chromosome becomes the visible bivalent
during prophase I of meiosis. So, an organism with 14 chromosomes will
have 14 bivalents.
7.
Note: The octad illustrated in (a) happens to have no crossover in the region
between the centromere and the gene being followed. A crossover here would
change the pattern of ascospores in the ascus but would not effect the 1:1 ratio
of alleles in the meiotic products.
8.
The four stages of mitosis are: prophase, metaphase, anaphase, and telophase.
The first letters, PMAT, can be remembered by a mnemonic such as: Playful
Mice Analyze Twice.
The five stages of prophase I are: leptotene, zygotene, pachytene, diplotene, and
diakinesis. The first letters, LZPDD, can be remembered by a mnemonic such
as: Large Zoos Provide Dangerous Distractions.
Chapter Three 43
9.
Each diploid meiocyte undergoes meiosis to form four products and then each
undergoes a postmeiotic mitotic division to give a total of eight ascospores per
meiocyte. In total then, 100 meiocytes will give rise to 800 ascospores.
10. The resulting cells will have the identical genotype as the original cell: A/a ;
B/b.
11. Yes. It could work but certain DNA repair mechanisms (such as postreplication
recombination repair) could not be invoked prior to cell division. There would
be just two cells as products of this meiosis, rather than four.
12. The general formula for the number of different male/female centromeric
combinations possible is 2n, where n = number of different chromosome pairs.
In this case, 25 = 32.
13. Because the DNA levels vary six-fold, the range covers cells that are haploid
(spores or cells of the gametophyte stage) to cells that are triploid (the
endosperm) and dividing (after DNA has replicated but prior to cell division).
The following cells would fit the DNA measurements:
0.7
1.4
2.1
2.8
4.2
14.
haploid cells
diploid cells in G1 or haploid cells after S but prior to cell division
triploid cells of the endosperm
diploid cells after S but prior to cell division
triploid cells after S but prior to cell division
Parent cell
a+
a+
b
a+
a+
b
b
Chromosome duplication
b
a+
a+
b
b
a+
b
Segregation
Daughter cells
15. PFGE separates DNA molecules by size. When DNA is carefully isolated from
Neurospora (which has seven different chromosomes) seven bands should be
produced using this technique. Similarly, the pea has seven different
44
Chapter Three
chromosomes and will produce seven bands (homologous chromosomes will
co-migrate as a single band).
16. There is a total of 4 m of DNA and nine chromosomes per haploid set. On
average, each is 4/9 m long. At metaphase, their average length is 13 µm, so the
average packing ratio is 13  10–6 m : 4.4  10–1 m or roughly 1:34,000! This
remarkable achievement is accomplished through the interaction of the DNA
with proteins. At its most basic, eukaryotic DNA is associated with histones in
units called nucleosomes and during mitosis, coils into a solenoid. As loops, it
associates with and winds into a central core of nonhistone protein called the
scaffold.
17. The nucleus contains the genome and separates it from the cytoplasm. However,
during cell division, the nuclear envelope dissociates (breaks down). It is the job
of the microtubule-based spindle to actually separate the chromosomes (divide
the genetic material) around which nuclei reform during telophase. In this sense,
it can be viewed as a passive structure that is divided by the cell’s cytoskeleton.
18. Yes. Half of our genetic makeup is derived from each parent, each parent’s
genetic makeup is derived half from each of their parents, etc.
19. Because the “half” inherited is very random, the chances of receiving exactly the
same half is vanishingly small. Ignoring recombination and focusing just on
which chromosomes are inherited from one parent (for example, the one they
inherited from their father or the one from their mother?), there are 223 =
8,388,608 possible combinations!
20. Remember, the endosperm is formed from two polar nuclei (which are
genetically identical) and one sperm nucleus.
Female
s/s
S/S
S/s
Male
S/S
s/s
S/s
Polar nuclei
s and s
S and S
1/ S and S
2
1/
21. a.
2
s and s
Sperm
S
s
1/ S
2
1/
2
s
Endosperm
S/s/s
S/S/s
1/ S/S/S
4
1/ S/S/s
4
1/ S/s/s
4
1/ s/s/s
4
A sporophyte of A/a ; B/b genotype will produce gametophytes in the
following proportions:
1/ A ; B
4
1/ A ; b
4
1/ a ; B
4
1/ a ; b
4
Chapter Three 45
b.
Random fertilization of the spores from the above gametophytes can occur
4  4 = 16 possible ways. Four of these combinations (A ; B a ; b, a ; b A
; B, A ; b a ; B, a ; BA ; b) will result in the desired A/a ; B/b sporophyte
genotype. Therefore 1/4 of next generation should be of this genotype.
22. All daughter cells will still be A/a ; B/b ; C/c. Mitosis generates daughter cells
genetically identical with that of the original cell.
23.
ad– ; a  ad+ ; 
P
Transient diploid
F1
24.
ad+/ad– ; a/
1/
ad+ ; a, white
1/ ad– ; a, purple
4
1/ ad+ ; , white
4
1/ ad– ; , purple
4
4
fern
Mitosis
sporophyte
gametophyte
Meiosis
sporophyte (sporangium)
moss
sporophyte
sporophyte (atheridium and
archegonium)
gametophyte
25.
plant
sporophyte
gametophyte
sporophyte (anther and ovule)
pine tree
sporophyte
gametophyte
sporophyte (pine cone)
mushroom
sporophyte
gametophyte
sporophyte (ascus or basidium)
frog
somatic cells
gonads
butterfly
somatic cells
gonads
snail
somatic cells
gonads
This problem is tricky because the answers depend on how a cell is defined. In
general, geneticists consider the transition from one cell to two cells to occur
with the onset of anaphase in both mitosis and meiosis, even though cytoplasmic
division occurs at a later stage.
a.
46 chromosomes, each with two chromatids = 92 chromatids
46
Chapter Three
b.
46 chromosomes, each with two chromatids = 92 chromatids
c.
46 physically separate chromosomes in each of two about-to-be-formed
cells
d.
23 chromosomes in each of two about-to-be-formed cells, each with two
chromatids = 46 chromatids
e.
23 chromosomes in each of two about-to-be-formed cells
26. (5) chromosome pairing (synapsis)
27. His children will have to inherit the satellite-containing 4 (probability = 1/2), the
abnormally-staining 7 (probability = 1/2), and the Y chromosome (probability =
1/ ). To get all three, the probability is (1/ )(1/ )(1/ ) = 1/ .
2
2
2
2
8
28. The parental set of centromeres can match either parent, which means there are
two ways to satisfy the problem. For any one pair, the probability of a
centromere from one parent going into a specific gamete is 1/2. For n pairs, the
probability of all the centromeres being from one parent is (1/2)n. Therefore, the
total probability of having a haploid complement of centromeres from either
parent is 2(1/2)n = (1/2)n–1.
29. Dear Monk Mendel:
I have recently read your most engrossing manuscript detailing the results of
your most wise experiments with garden peas. I salute both your curiosity and
your ingenuity in conducting said experiments, thereby opening up for scientific
exploration an entire new area of our Maker’s universe. Dear Sir, your findings
are extraordinary!
While I do not pretend to compare myself to you in any fashion, I beg to bring to
your attention certain findings I have made with the aid of that most fascinating
and revealing instrument, the microscope. I have been turning my attention to
the smallest of worlds with an instrument that I myself have built, and I have
noticed some structures that may parallel in behavior the factors which you have
postulated in the pea.
I have worked with grasshoppers, however, not your garden peas. Although you
are a man of the cloth, you are also a man of science, and I pray that you will not
be offended when I state that I have specifically studied the reproductive organs
of male grasshoppers. Indeed, I did not limit myself to studying the organs
themselves; instead, I also studied the smaller units that make up the male
organs and have beheld structures most amazing within them.
Chapter Three 47
These structures are contained within numerous small bags within the male
organs. Each bag has a number of these structures, which are long and threadlike
at some times and short and compact at other times. They come together in the
middle of a bag, and then they appear to divide equally. Shortly thereafter, the
bag itself divides, and what looks like half of the threadlike structures goes into
each new bag. Could it be, Sir, that these threadlike structures are the very same
as your factors? I know, of course, that garden peas do not have male organs in
the same way that grasshoppers do, but it seems to me that you found it
necessary to emasculate the garden peas in order to do some crosses, so I do not
think it too far-fetched to postulate a similarity between grasshoppers and
garden peas in this respect.
Pray, Sir, do not laugh at me and dismiss my thoughts on this subject even
though I have neither your excellent training nor your astounding wisdom in the
Sciences. I remain your humble servant to eternity!
30. When results of a cross are sex-specific, sex linkage should be considered. In
moths, the heterogametic sex is actually the female while the male is the
homogametic sex. Assuming that dark (D) is dominant to light (d), then the data
can be explained by the dark male being heterozygous (D/d) and the dark female
being hemizygous (D). All male progeny will inherit the D allele from their
mother and therefore be dark while half the females will inherit D from their
father (and be dark) and half will inherit d (and be light).
31. a.
Both the gametophyte and the sporophyte are closer in shape to the mother
than the father. Note that a size increase occurs in each type of cross.
b.
Gametophyte and sporophyte morphology are affected by extranuclear
factors. Leaf size may be a function of the interplay between nuclear
genome contributions.
c.
If extranuclear factors are affecting morphology while nuclear factors are
affecting leaf size, then repeated backcrosses could be conducted, using the
hybrid as the female. This would result in the cytoplasmic information
remaining constant while the nuclear information becomes increasingly like
that of the backcross parent. Leaf morphology should therefore remain
constant while leaf size would decrease toward the size of the backcross
parent.
32. The goal here is to generate a plant with the cytoplasm of plant A and the
nuclear genome predominantly of plant B. Remember that the cytoplasm is
contributed by the egg only. So using plant A as the maternal parent, cross to B
(as the paternal parent) and then backcross the progeny of this cross using plant
B again as the paternal parent. Repeat for several generations until virtually the
entire nuclear genome is from the B parent.
48
Chapter Three
33. If the variegation is due to a chloroplast mutation, then the phenotype of the
offspring will be controlled solely by the phenotype of the maternal parent. Look
for flowers on white branches and test to see if they produce seeds that grow
into all white plants regardless of the source of the pollen. To test whether a
dominant nuclear mutation is responsible for the variegation, cross pollen from
flowers on white branches to green plants (or flowers on green branches of same
plant) and see if half the progeny are white and half are green (assuming the
dominant mutation is heterozygous) or all white, if the dominant mutation is
homozygous.
34. Progeny plants inherited only normal cpDNA (lane 1); only mutant cpDNA
(lane 2); or both (lane 3). In order to get homozygous cpDNA, seen in lanes 1
and 2, segregation of chloroplasts had to occur.
Chapter Three 49
CHALLENGING PROBLEMS
35. In the following schematic drawings, chromosomes (or chromatids) that are
radioactive are indicated be the grains that would be observed after
radioautography. After the second mitotic division, a number of outcomes are
possible due to the random alignment and separation of the radioactive and nonradioactive chromatids.
*
* *
** *
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*** *
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****
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Prophase of
first mitosis
Telophase of
first mitosis
*
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Prophase of
second mitosis
*
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*
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*
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*
*
** *
*
* ***
*
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*
Chapter Three
*
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Probability
1/8
and
OR
*
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and
*
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OR
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and
1/4
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and
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Telophase of
second mitosis
*
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*
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50
1/8
Chapter Three 51
36.
37. a.
In a diploid cell, expect two chromosomes (a pair of homologs) to each
have a single locus of radioactivity.
b.
Expect many regions of radioactivity scattered throughout the
chromosomes. The exact number and pattern would be dependent on the
specific sequence in question, and where and how often it is present within
the genome.
c.
The multiple copies of the genes for ribosomal RNA are organized into
large tandem arrays called nucleolar organizers (NO). Therefore, expect
broader areas of radioactivity compared to (a). The number of these regions
would equal the number of NO present in the organism.
d.
Each chromosome end would be labeled by telomeric DNA.
e.
The multiple repeats of this heterochromatic DNA are organized into large
tandem arrays. Therefore, expect broader areas of radioactivity compared to
(a). Also, there may be more than one area in the genome of the same
simple repeat.
52
Chapter Three
38. The following is meant to be examples of what is possible. It is also possible,
for instance, that more than one band would be present in (a), depending on the
position of the restriction sites within the sequence complementary to the probe
used.
a
b
c
For (d) and (e), the specifics or where the DNA is cut relative to the telomeric
DNA or heterochromatic DNA will effect what is observed. Assuming the
restriction sites are not within the telomeric or heterochromatic DNA, then (d)
will be similar to (b), and (c) will have one or several very large bands.
39. First, examine the crosses and the resulting genotypes of the endosperm:
Female
Male
Polar nuclei
Sperm
Endosperm
ƒ´/ƒ´
(floury)
ƒ´´/ƒ´´
ƒ´ and ƒ´
ƒ´´/ƒ´´
ƒ´/ƒ´/ƒ´´
ƒ´´/ƒ´´
(flinty)
ƒ´/ƒ´
ƒ´´ and ƒ´´
ƒ´/ƒ´
ƒ´´/ƒ´´/ƒ´
As can be seen, the phenotype of the endosperm correlates to the predominant
allele present.
40.
(1)
(2)
Impossible: the alleles of the same genes are on nonhomologous
chromosomes
Meiosis II
Chapter Three 53
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
Meiosis II
Meiosis II
Mitosis
Impossible: appears to be mitotic anaphase but alleles of sister
chromatids are not identical
Impossible: too many chromosomes
Impossible: too many chromosomes
Impossible: too many chromosomes
Meiosis I
Impossible: appears to be meiosis of homozygous a/a ; B/B
Impossible: the alleles of the same genes are on nonhomologous
chromosomes
41. Recall that each cell has many mitochondria, each with numerous genomes.
Also recall that cytoplasmic segregation is routinely found in mitochondrial
mixtures within the same cell.
The best explanation for this pedigree is that the mother in generation I
experienced a mutation in a single cell that was a progenitor of her egg cells
(primordial germ cell). By chance alone, the two males with the disorder in the
second generation were from egg cells that had experienced a great deal of
cytoplasmic segregation prior to fertilization, while the two females in that
generation received a mixture.
The spontaneous abortions that occurred for the first woman in generation II
were the result of extensive cytoplasmic segregation in her primordial germ
cells: aberrant mitochondria were retained. The spontaneous abortions of the
second woman in generation II also came from such cells. The normal children
of this woman were the result of extensive segregation in the opposite direction:
normal mitochondria were retained. The affected children of this woman were
from egg cells that had undergone less cytoplasmic segregation by the time of
fertilization, so that they developed to term but still suffered from the disease.
42. For the following, S will signify cytoplasm of a male-sterile line and N will
signify cytoplasm of a non-male-sterile line. Rf will signify the dominant nuclear
restorer allele and rf, the recessive non-restorer allele.
a.
If S rf/rf (male-sterile plants) are crossed with pollen from N Rf/Rf plants,
the offspring will all be S Rf/rf and male fertile. If these offspring are then
crossed with pollen from N rf/rf plants, half the offspring will be S Rf/rf
(male-fertile) and half will be S rf/rf (male-sterile). The S cytoplasm will
not be altered or affected even though the maternal offspring parent plant
was Rf/rf.
b.
The cross is S rf/rf  N Rf/Rf so all the progeny will be S Rf/rf and malefertile.
54
Chapter Three
c.
The cross is S Rf/rf  N rf/rf so half the progeny will be S Rf/rf (malefertile) and half will be S rf/rf (male-sterile).
d.
i.
The cross is S rf-1/rf-1 ; rf-2/rf-2  N Rf-1/rf-1 ; Rf-2/rf-2
The progeny will be: 1/4 S Rf-1/rf-1 ; Rf-2/rf-2 (male-fertile)
1/ S Rf-1/rf-1 ; rf-2/rf-2 (male-fertile)
4
1/ S rf-1/rf-1 ; Rf-2/rf-2 (male-fertile)
4
1/ S rf-1/rf-1 ; rf-2/rf-2 (male-sterile)
4
ii.
The cross is S rf-1/rf-1 ; rf-2/rf-2  N Rf-1/Rf-1 ; rf-2/rf-2
The progeny will all be: S Rf-1/rf-1 ; rf-2/rf-2 (male-fertile)
iii. The cross is S rf-1/rf-1 ; rf-2/rf-2  N Rf-1/rf-1 ; rf-2/rf-2
The progeny will be: 1/2 S Rf-1/rf-1 ; rf-2/rf-2 (male-fertile)
1/ S rf-1/rf-1 ; rf-2/rf-2 (male-sterile)
2
vi. The cross is S rf-1/rf-1 ; rf-2/rf-2  N Rf-1/rf-1 ; Rf-2/Rf-2
The progeny will be: 1/2 S Rf-1/rf-1 ; Rf-2/rf-2 (male-fertile)
1/ S rf-1/rf-1 ; Rf-2/rf-2 (male-fertile)
2