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ISBN 10: 1-292-03922-1
ISBN 10: 1-269-37450-8
ISBN 13: 978-1-292-03922-0
ISBN 13: 978-1-269-37450-7
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Li n ka g e a nd C hr omos ome ma pp ing in eu k a ry ot es
InsIghts and solutIons
1. In rabbits, black color (B) is dominant to brown (b), while full
color (C) is dominant to chinchilla (c ch). The genes controlling
these traits are linked. Rabbits that are heterozygous for both traits
and express black, full color are crossed to rabbits that express
brown, chinchilla with the following results:
31
brown, chinchilla
34
black, full
16
brown, full
19
black, chinchilla
Solution: First, determine the arrangement of the alleles on the
homologs of the heterozygous crossover parent (the female in this
case). To do this, locate the most frequent reciprocal phenotypes,
which arise from the noncrossover gametes—these are phenotypes
(1) and (2). Each phenotype represents the arrangement of alleles
on one of the homologs. Therefore, the arrangement is
Determine the arrangement of alleles in the heterozygous parents
and the map distance between the two genes.
Solution: This is a two-point map problem, where the two most
prevalent reciprocal phenotypes are the noncrossovers. The less
frequent reciprocal phenotypes arise from a single crossover. The
arrangement of alleles is derived from the noncrossover phenotypes because they enter gametes intact.
The single crossovers give rise to 35>100 offspring (35 percent). Therefore, the distance between the two genes is 35 mu.
B
c ch
b
C
c ch
B
C
b
c ch
black, full
B
c ch
Noncrossovers
b
c ch
b
c ch
c ch
b
brown, chinchilla
Single crossovers
b
c ch
black, chinchilla
b
b
br
Ly
Sb
br
will yield Ly + br and + Sb + as phenotypes. Inspection shows
that these categories (5 and 6) are actually single crossovers, not
double crossovers. Therefore, the sequence is incorrect as written.
Only two other sequences are possible: The br gene is either to the
left of Ly (Case A), or it is between Ly and Sb (Case B).
C
brown, full
c ch
2. In Drosophila, Lyra (Ly) and Stubble (Sb) are dominant mutations located at locus 40 and 58, respectively, on chromosome III.
A recessive mutation with bright red eyes is discovered and shown
also to be located on chromosome III. A map is obtained by crossing a female who is heterozygous for all three mutations to a male
that is homozygous for the bright-red mutation (which we will call
br), and the data in the table are generated. Determine the location
of the br mutation on chromosome III.
Phenotype Sb
Second, find the correct sequence of the three loci along the
chromosome. This is done by determining which sequence yields
the observed double-crossover phenotypes, which are the least frequent reciprocal phenotypes (7 and 8). If the sequence is correct as
written, then the double crossover depicted here,
b
Ly
Number
(1) Ly
Sb
br
404
(2) +
+
+
422
(3) Ly
+
+
18
(4) +
Sb
br
16
(5) Ly
+
br
75
(6) +
Sb
+
59
(7) Ly
Sb
+
4
(8) +
+
br
2
Total
= 1000
br
Case A
Ly
Sb
Ly
Case B
br
Sb
Double crossovers
Sb
Ly
br
Double crossovers
Sb
and
and
Ly
br
Comparison with the actual data shows that Case B is correct.
The double-crossover gametes yield flies that express Ly and Sb
but not br, or express br but not Ly and Sb. Therefore, the correct
arrangement and sequence are as follows.
Ly
br
Sb
179
L i n k a g e a n d C h r o mo s om e ma pping in eu ka r yo te s
Once this sequence is found, determine the location of br
relative to Ly and Sb. A single crossover between Ly and br, as
shown here,
Ly
br
Sb
The final map shows that br is located at locus 44, since Lyra and
Stubble are known.
40
Ly
yields flies that are Ly + + and + br Sb (phenotypes 3 and 4).
Therefore, the distance between the Ly and br loci is equal to
(18 + 16 + 4 + 2)>1000 = 40>1000 = 0.04 = 4 mu
Remember to add the double crossovers because they represent two single crossovers occurring simultaneously. You need to
know the frequency of all crossovers between Ly and br, so they
must be included.
Similarly, the distance between the br and Sb loci is derived
mainly from single crossovers between them.
Ly
br
Sb
This event yields Ly br+ and + + Sb phenotypes (phenotypes 5
and 6). Therefore, the distance equals
(4)
44
(14)
58
br
Sb
3. In reference to Problem 2, a student predicted that the mutation was actually the known mutation scarlet located at locus 44.0.
Suggest an experimental cross that would confirm this prediction.
Solution: Since the scarlet locus is identical to the experimental
assignment, it is reasonable to hypothesize that the bright-red eye
mutation is an allele at the scarlet locus.
To test this hypothesis, you could perform complementation analysis by crossing females expressing the bright-red mutation with known scarlet males. If the two mutations are alleles,
no complementation will occur and all progeny will reveal a
bright-red mutant eye phenotype. If complementation occurs,
all progeny will express normal brick-red (wild-type) eyes, since
the bright-red mutation and scarlet are at different loci (they are
probably very close together). In such a case, all progeny will be
heterozygous at both the bright-red eye and the scarlet loci and
will not express either mutation because they are both recessive.
This type of complementation analysis is called an allelism test.
(75 + 59 + 4 + 2)>1000 = 140>1000 = 0.14 = 14 mu
probLEMs anD Discussion QuEstions
How Do we Know
?
1. In this chapter, we focused on linkage, chromosomal mapping,
and many associated phenomena. In the process, we found many
opportunities to consider the methods and reasoning by which
much of this information was acquired. What answers would
you propose to the following fundamental questions?
(a) How was it established experimentally that the frequency of
recombination (crossing over) between two genes is related
to the distance between them along the chromosome?
(b) How do we know that specific genes are linked on a single
chromosome, in contrast to being located on separate chromosomes?
(c) How do we know that crossing over results from a physical
exchange between chromatids?
(d) How do we know that sister chromatids undergo recombination during mitosis?
2. What is the significance of crossing over (which leads to genetic
recombination) to the process of evolution?
3. Describe the cytological observation that suggests that crossing
over occurs during the first meiotic prophase.
4. Why does more crossing over occur between two distantly linked
genes than between two genes that are very close together on the
same chromosome?
5. Why is a 50 percent recovery of single-crossover products the
upper limit, even when crossing over always occurs between two
linked genes?
180
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6. Why are double-crossover events expected less frequently than
single-crossover events?
7. What is the proposed basis for positive interference?
8. What three essential criteria must be met in order to execute a
successful mapping cross?
9. The genes dumpy wings (dp), clot eyes (cl), and apterous wings
(ap) are linked on chromosome II of Drosophila. In a series of
two-point mapping crosses, the genetic distances shown below
were determined. What is the sequence of the three genes?
d p–a p
42
d p–c l
3
a p–c l
39
10. Colored aleurone in the kernels of corn is due to the dominant
allele R. The recessive allele r, when homozygous, produces colorless aleurone. The plant color (not kernel color) is controlled
by another gene with two alleles, Y and y. The dominant Y allele
results in green color, whereas the homozygous presence of the
recessive y allele causes the plant to appear yellow. In a testcross
between a plant of unknown genotype and phenotype and a
plant that is homozygous recessive for both traits, the following
progeny were obtained:
colored, green
88
colored, yellow
12
colorless, green
8
colorless, yellow
92
Li n ka g e a nd C hr omos ome ma pp ing in eu k a ry ot es
Explain how these results were obtained by determining the exact genotype and phenotype of the unknown plant, including the
precise association of the two genes on the homologs (i.e., the
arrangement).
11. In the cross shown here, involving two linked genes, ebony (e) and
claret (ca), in Drosophila, where crossing over does not occur in
males, offspring were produced in a (2 + :1 ca :1 e) phenotypic ratio:
+
e
{
ca+
e
determination of sex was made in the F 2 data. (a) Using proper
nomenclature, determine the genotypes of the P 1 and F 1 parents.
(b) Determine the sequence of the three genes and the map distance between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence;
does this represent positive or negative interference?
Phenotype
sc
+
+
sc
sc
+
sc
+
ca +
*
e + ca
e + ca
These genes are 30 mu apart on chromosome III. What did
crossing over in the female contribute to these phenotypes?
12. In a series of two-point map crosses involving five genes located
on chromosome II in Drosophila, the following recombinant
(single-crossover) frequencies were observed:
pr-adp
29
pr-vg
13
pr-c
21
pr-b
6
adp-b
35
adp-c
8
adp-vg
16
vg-b
19
vg-c
8
c-b
27
Female A
Female B
d b +
+ + c
d + +
+ b c
(1) d b c
v
+
v
+
v
+
+
v
314
280
150
156
46
30
10
14
15. A cross in Drosophila involved the recessive, X-linked genes
yellow body (y), white eyes (w), and cut wings (ct). A yellowbodied, white-eyed female with normal wings was crossed to a
male whose eyes and body were normal, but whose wings were
cut. The F 1 females were wild type for all three traits, while the F 1
males expressed the yellow-body, white-eye traits. The cross was
carried to F 2 progeny, and only male offspring were tallied. On
the basis of the data shown here, a genetic map was constructed.
(a) Diagram the genotypes of the F 1 parents. (b) Construct a
map, assuming that w is at locus 1.5 on the X chromosome. (c)
Were any double-crossover offspring expected? (d) Could the F 2
female offspring be used to construct the map? Why or why not?
(a) If the adp gene is present near the end of chromosome II
(locus 83), construct a map of these genes.
(b) In another set of experiments, a sixth gene (d) was tested
against b and pr, and the results were d - b = 17% and
d - pr = 23%. Predict the results of two-point maps between d and c, d and vg, and d and adp.
13. Two different female Drosophila were isolated, each heterozygous for the autosomally linked genes black body (b), dachs tarsus (d), and curved wings (c). These genes are in the order d–b–c,
with b closer to d than to c. Shown in the following table is the
genotypic arrangement for each female, along with the various
gametes formed by both. Identify which categories are noncrossovers (NCO), single crossovers (SCO), and double crossovers
(DCO) in each case. Then, indicate the relative frequency with
which each will be produced.
T
Offspring
s
+
s
+
+
s
s
+
Gamete formation
T
(5) d + +
(1) d b +
(5) d b c
(2) + + +
(6) + b c
(2) + + c
(6) + + +
(3) + + c
(7) d + c
(3) d + c
(7) d + +
(4) d b +
(8) + b +
(4) + b +
(8) + b c
14. In Drosophila, a cross was made between females expressing the
three X-linked recessive traits, scute bristles (sc), sable body (s),
and vermilion eyes (v), and wild-type males. All females were wild
type in the F 1, while all males expressed all three mutant traits.
The cross was carried to the F 2 generation and 1000 offspring
were counted, with the results shown in the following table. No
Phenotype
y
+
y
+
+
y
y
+
+
w
w
+
+
w
+
w
Male Offspring
ct
+
ct
+
ct
+
+
ct
9
6
90
95
424
376
0
0
16. In Drosophila, Dichaete (D) is a mutation on chromosome III
with a dominant effect on wing shape. It is lethal when homozygous. The genes ebony body (e) and pink eye (p) are recessive
mutations on chromosome III. Flies from a Dichaete stock were
crossed to homozygous ebony, pink flies, and the F 1 progeny
with a Dichaete phenotype were backcrossed to the ebony, pink
homozygotes. Using the results of this backcross shown in the
following table, (a) diagram the cross, showing the genotypes
of the parents and offspring of both crosses. (b) What is the sequence and interlocus distance between these three genes?
Phenotype
Dichaete
ebony, pink
Dichaete, ebony
pink
Dichaete, pink
ebony
Dichaete, ebony, pink
wild type
Number
401
389
84
96
2
3
12
13
181
L i n k a g e a n d C h r o mo s om e ma pping in eu ka r yo te s
17. Drosophila females homozygous for the third chromosomal
genes pink eye (p) and ebony body (e) were crossed with males
homozygous for the second chromosomal gene dumpy wings
(dp). Because these genes are recessive, all offspring were wild
type (normal). F 1 females were testcrossed to triply recessive
males. If we assume that the two linked genes (p and e) are 20 mu
apart, predict the results of this cross. If the reciprocal cross were
made (F 1 males—where no crossing over occurs—with triply recessive females), how would the results vary, if at all?
18. In Drosophila, the two mutations Stubble bristles (Sb) and curled
wings (cu) are linked on chromosome III. Sb is a dominant gene
that is lethal in a homozygous state, and cu is a recessive gene. If
a female of the genotype
Sb cu
+
+
is to be mated to detect recombinants among her offspring, what
male genotype would you choose as her mate?
19. A female of genotype
a b c
+ + +
produces 100 meiotic tetrads. Of these, 68 show no crossover
events. Of the remaining 32, 20 show a crossover between a and b,
10 show a crossover between b and c, and 2 show a double crossover between a and b and between b and c. Of the 400 gametes
produced, how many of each of the eight different genotypes will
be produced? Assuming the order a–b–c and the allele arrangement shown above, what is the map distance between these loci?
20. In a plant, fruit color is either red or yellow, and fruit shape is
either oval or long. Red and oval are the dominant traits. Two
plants, both heterozygous for these traits, were testcrossed, with
the results shown in the following table. Determine the location
of the genes relative to one another and the genotypes of the two
parental plants.
Progeny
Phenotype
red, long
yellow, oval
red, oval
yellow, long
Total
Plant A
Plant B
46
44
5
5
4
6
43
47
100
100
21. Drosophila melanogaster has one pair of sex chromosomes (XX
or XY) and three autosomes (chromosomes II, III, and IV). A
genetics student discovered a male fly with very short (sh) legs.
Using this male, the student was able to establish a pure-breeding
stock of this mutant and found that it was recessive. She then
incorporated the mutant into a stock containing the recessive
gene black (b, body color, located on chromosome II) and the
recessive gene pink (p, eye color, located on chromosome III). A
female from the homozygous black, pink, short stock was then
mated to a wild-type male. The F 1 males of this cross were all
wild type and were then backcrossed to the homozygous b, p, sh
females. The F 2 results appeared as shown in the following table,
and no other phenotypes were observed. (a) Based on these results, the student was able to assign sh to a linkage group (a chromosome). Determine which chromosome, and include step-bystep reasoning. (b) The student repeated the experiment, making
the reciprocal cross: F 1 females backcrossed to homozygous b,
p, sh males. She observed that 85 percent of the offspring fell
into the given classes, but that 15 percent of the offspring were
equally divided among b+ p, b+ +, +shp, and + sh + phenotypic
182
males and females. How can these results be explained, and what
information can be derived from these data?
Phenotype
wild
pink*
black, short
black, pink, short
Female
Male
63
58
55
69
59
65
51
60
*
Pink indicates that the other two traits are wild type (normal).
Similarly, black, short offspring are wild type for eye color.
22. In Drosophila, a female fly is heterozygous for three mutations:
Bar eyes (B), miniature wings (m), and ebony body (e). (Note
that Bar is a dominant mutation.) The fly is crossed to a male
with normal eyes, miniature wings, and ebony body. The results
of the cross are shown in the following list. Interpret the results
of this cross. If you conclude that linkage is involved between any
of the genes, determine the map distance(s) between them.
miniature
111
wild type
29
Bar, ebony
101
Bar, miniature, ebony
31
Bar
117
Bar, miniature
26
ebony
35
miniature, ebony
115
23. An organism of the genotype AaBbCc was testcrossed to a triply
recessive organism (aabbcc). The genotypes of the progeny are in
the following table.
AaBbCc
20
AaBbcc
20
aabbCc
20
aabbcc
20
AabbCc
5
Aabbcc
5
aaBbCc
5
aaBbcc
5
(a) Assuming simple dominance and recessiveness in each gene
pair, if these three genes were all assorting independently,
how many genotypic and phenotypic classes would result in
the offspring, and in what proportion?
(b) Answer part (a) again, assuming the three genes are so tightly
linked on a single chromosome that no crossover gametes
were recovered in the sample of offspring.
(c) What can you conclude from the actual data about the location of the three genes in relation to one another?
24. Based on our discussion of the potential inaccuracy of mapping
(see Figure 12), would you revise your answer to Problem 23(c)?
If so, how?
25. In Creighton and McClintock’s experiment demonstrating that
crossing over involves physical exchange between chromosomes
(see Section 6), explain the importance of the cytological markers (the translocated segment and the chromosome knob) in the
experimental rationale.
26. Traditional gene mapping has been applied successfully to a variety of organisms including yeast, fungi, maize, and Drosophila.
However, human gene mapping has only recently shared a similar spotlight. What factors have delayed the application of traditional gene-mapping techniques in humans?
27. DNA markers have greatly enhanced the mapping of genes in
humans. What are DNA markers, and what advantage do they
confer?
28. Are sister chromatid exchanges effective in producing genetic
variability in an individual? in the offspring of individuals?
Li n ka g e a nd C hr omos ome ma pp ing in eu k a ry ot es
soLutions to sELEctED probLEMs anD Discussion QuEstions
answers to now solve this
1. (a) 1/4 AaBb 1/4 Aabb 1/4 aaBb 1/4 aabb
(b) 1/4 AaBb 1/4 Aabb 1/4 aaBb 1/4 aabb
(c) If the arrangement is AB/ab * ab/ab
then the two types of offspring will be as follows:
1/2 Ab/ab
1/2 aB/ab
If, however, A and B are not coupled, then the symbolism
would be Ab/aB * aabb.
The offspring would occur as follows:
1/2 Ab/ab
1/2 aB/ab
2. The most frequent classes are PZ and pz. These classes represent
the parental (noncrossover) groups, which indicates that the original parental arrangement in the testcross was PZ/pz * pz/pz.
Adding the crossover percentages together (6.9 + 7.1) gives 14
percent, which would be the map distance between the two genes.
3. Examine the progeny list to see which types are not present. In this
case, the double crossover classes are the following: + + c and a b +
(a, b) Gene b is in the middle and the arrangement is as follows.
+ b c/ a + +
a - b = 7 map units
b - c = 2 map units
(c) The progeny phenotypes that are missing are + + c and a b
+, of which, from 1000 offspring, 1.4 (0.07 * 0.02 * 1000)
would be expected. Perhaps by chance or some other unknown selective factor, they were not observed.
solutions to problems and Discussion Questions
4. With some qualification, especially around the centromeres
and telomeres, one can say that crossing over is somewhat randomly distributed over the length of the chromosome. Two loci
that are far apart are more likely to have a crossover between
them than two loci that are close together.
6. If the probability of one event is 1/X, the probability of two
events occurring at the same time will be 1/X2.
10. The heterozygous parent in the test cross is RY/ry * ry/ry with
the two dominant alleles on one chromosome and the two recessives on the homolog. The map distance would be 10 map
units between the R and Y loci.
12. The map for parts (a) and (b) is the following:
d............b.........pr..........vg........c.........adp
31
48
54
67
75
83
(+++++)+++++*
Map units
The expected map units between d and c would be 44, between
d and vg 36, and between d and adp 52. However, because
there is a theoretical maximum of 50 map units possible between two loci in any one cross, that distance would be below
the 52 determined by simple subtraction.
14. (a) P1: sc s v/sc s v * + + +/Y
F1: + + +/sc s v * sc s v/Y
(b) The map distances are determined by first writing the
proper arrangement and sequence of genes, and then computing the distances between each set of genes.
sc v s
+ + +
sc - v = 33 percent (map units)
v - s = 10 percent (map units)
(c, d) The coefficient of coincidence = 0.727, which indicates
that there were fewer double crossovers than expected;
therefore, positive chromosomal interference is present.
16. (a) Represent the Dichaete gene as an uppercase letter because it is dominant. At this point, the gene sequence is not
given.
P1: D + +/ + + + * + e p/+ e p
F1: D + +/+ e p * + e p/+ e p
F2:
(b) F1:
D + +/+ e p
+ e p/+ e p
D e +/+ e p
+ + p/+ e p
D + p/+ e p
+ e +/+ e p
D e p/+ e p
+ + +/+ e p
Dichaete
ebony, pink
Dichaete, ebony
pink
Dichaete, pink
ebony
Dichaete, ebony, pink
wild type
D + +/+ p e * + p e/+ p e
D - p = 3.0 map units
p - e = 18.5 map units
18. One would use the typical testcross arrangement with the
curled gene so the male parent arrangement would be +
cu/ + cu.
20. Assign the following symbols, for example:
R = Red
O = Oval
r = yellow
o = long
Progeny A: Ro/rO * rroo = 10 map units
Progeny B: RO/ro * rroo = 10 map units
22. Begin with a set of symbols as indicated below:
B+ = wild eye shape
m+ = wild wings
e+ = wild body color
B = Bar eye shape
m = miniature wings
e = ebony body color
Superficially, the cross would be as follows:
B+B m+m e+e * B+? m? e
(The ? is used at this point to indicate that we have no information allowing us to decide whether any of the alleles
in the male are X-linked.) The arrangement is as indicated
in the following: B m+/B+m; e+/e
Notice that a semicolon is used to indicate that the ebony locus
is on a different chromosome. At this point and without prior
knowledge, we still don’t know whether any of the genes are
X-linked; however, it is of no consequence to the solution of the
problem. (In actuality, both B and m loci are X-linked.) To determine the map distances (again, ebony is out of the mapping
picture at this point because it is not linked to either B or m):
111
117
26
29
+
+
+
+
115
101
31
35
=
=
=
=
226
218
57
64
=
=
=
=
parental
parental
crossover
crossover
Mapping the distance between B and m would be as follows:
(57 + 64)/(226 + 218 + 57 + 64) * 100 =
121/565 * 100 = 21.4 map units
183