Chapter 17
21. MM N OO would be classified as 2n − 1; MM NN OO would be classified as 2n ; and MMM NN PP would be classified as 2n + 1.
24. There would be one possible quadrivalent.
27. Seven chromosomes.
29. Cells destined to become pollen grains can be induced by cold treatment to grow into embryoids. These embryoids can then be grown on agar to form monoploid plantlets.
31. Yes.
34. No.
36. An acentric fragment cannot be aligned or moved in meiosis (or mitosis) and is consequently lost.
39. Very large deletions tend to be lethal, likely owing to genomic imbalance or the unmasking of recessive lethal genes. Therefore, the observed very large pairing loop is more likely to be from a heterozygous inversion.
41. Williams syndrome is the result of a deletion of the 7q11.23 region of chromosome 7. Cri du chat syndrome is the result of a deletion of a significant part of the short arm of chromosome 5 (specifically bands 5p15.2 and 5p15.3). Both Turner syndrome (XO) and Down syndrome (trisomy 21) result from meiotic nondisjunction. The term syndrome is used to describe a set of phenotypes (often complex and varied) that are generally present together.
46. The order is b a c e d f.
|
Allele |
Band |
|
b |
1 |
|
a |
2 |
|
c |
3 |
|
e |
4 |
|
d |
5 |
|
f |
6 |
47. The data suggest that one or both breakpoints of the inversion are located within an essential gene, causing a recessive lethal mutation.
50.
When crossed with yellow females, the results would be
Xe/Ye+
gray males
Xe/Xe
yellow females
If the e+ allele was translocated to an autosome, the progeny would be as follows, where “A” indicates autosome:
P
Ae+/A; Xe/Y × A/A; Xe/Xe
F1
Ae+/A; Xe/Xe
gray female
Ae+/A; Xe/Y
gray male
A/A; Xe/Xe
yellow female
A/A; Xe/Y
yellow male
52.
|
Klinefelter syndrome |
XXY male |
|
Down syndrome |
trisomy 21 |
|
Turner syndrome |
XO female |
56.
If a hexaploid were crossed with a tetraploid, the result would be pentaploid.
Cross A/A with a/a/a/a to obtain A/a/a.
The easiest way is to expose the A/a* plant cells to colchicine for one cell division, which will result in a doubling of chromosomes to yield A/A/a*/a*.
Cross a hexaploid (a/a/a/a/a/a) with a diploid (A/A) to obtain A/a/a/a.
58.
The ratio of normal-
leaved to potato- leaved plants will be 5 : 1. If the gene is not on chromosome 6, there should be a 1 : 1 ratio of normal-
leaved to potato- leaved plants.
62.
The aberrant plant is semisterile, which suggests an inversion. Because the d-
f and y-p frequencies of recombination in the aberrant plant are normal, the inversion must implicate b through x.To obtain recombinant progeny when there has been an inversion requires the occurrence of either a double crossover within the inverted region or single crossovers between f and the inversion, which occurred someplace between f and b.
64. The original plant is homozygous for a translocation between chromosomes 1 and 5, with break points very close to genes P and S. Because of the close linkage, a ratio suggesting a monohybrid cross, instead of a dihybrid cross, was observed, both with selfing and with a testcross. All gametes are fertile because of homozygosity.
|
original plant: |
P S/p s |
|
tester: |
p s/p s |
F1 progeny: heterozygous for the translocation:
The easiest way to test this hypothesis is to look at the chromosomes of heterozygotes in meiosis I.
70. The original parents must have had the following chromosome constitution:
|
G. hirsutum |
26 large, 26 small |
|
G. thurberi |
26 small |
|
G. herbaceum |
26 large |
G. hirsutum is a polyploid derivative of a cross between the two Old World species, which could easily be checked by looking at the chromosomes.
72.
Loss of one X in the developing fetus after the two-
cell stage. Nondisjunction leading to Klinefelter syndrome (XXY), followed by a nondisjunctive event in one cell for the Y chromosome after the two-
cell stage, resulting in XX and XXYY. Nondisjunction of the X at the one-
cell stage. Fused XX and XY zygotes (from the separate fertilizations either of two eggs or of an egg and a polar body by one X-
bearing and one Y- bearing sperm). Nondisjunction of the X at the two-
cell stage or later.
75.
a. Each mutant is crossed with wild type, or
m × m+
The resulting tetrads (octads) show 1 : 1 segregation, indicating that each mutant is the result of a mutation in a single gene.
b. The results from crossing the two mutant strains indicate either that both strains are mutant for the same gene:
m1 × m2
or that they are mutant in different but closely linked genes:
m1 m2+ × m1+ m2
c. and d. Because phenotypically black offspring can result from nondisjunction (notice that, in cases C and D, black appears in conjunction with aborted spores), mutant 1 and mutant 2 are likely to be mutant in different but closely linked genes. The cross is therefore
m1 m2+ × m1+ m2
Case A is an NPD tetrad and would be the result of a four-
|
m1+ m2+ |
black |
|
m1+ m2+ |
black |
|
m1 m2 |
fawn |
|
m1 m2 |
fawn |
Case B is a tetratype and would be the result of a single crossover between one of the genes and the centromere.
|
m1+ m2+ |
black |
|
m1+ m2 |
fawn |
|
m1 m2+ |
fawn |
|
m1 m2 |
fawn |
Case C is the result of nondisjunction in meiosis I.
|
m1+ m2+ ; m1+ m2+ |
black |
|
m1+ m2+ ; m1+ m2+ |
black |
|
no chromosome |
abort |
|
no chromosome |
abort |
Case D is the result of recombination between one of the genes and the centromere followed by nondisjunction in meiosis II. For example,
|
m1+m2 ; m1m2+ |
black |
|
no chromosome |
abort |
|
m1m2+ |
fawn |
|
m1+m2 |
fawn |