1.
Prove
Mendel's Law of Segregation
2.
Prove
Mendel's Law of Independent Assortment.
3.
Suppose
you learned that "shmoos" may have long, oval, or round bodies and
that matings of "shmoos" gave the following offspring:
long X oval 52 long : 48 oval
long X round 99 oval
oval X round 51 oval : 50
round
oval X oval 24 long :
53 oval : 27 round
What hypothesis about the inheritance of "shmoo" shape would
be consistent with these results?
4.
Color‑blindness
is a sex‑linked trait in humans.
(a)
A
normal woman whose father was color‑blind marries a normal man. What will
be the proportion of color‑blind children?
(b)
How
could you ever get a color‑blind woman?
(c)
What
name is given to such a person as the female referenced in problem (b) above?
5.
Probability
and Genetic Events
(a)
What
are the chances that a family with 4 children will have 2 boys and 2 girls?
(b)
What
are the chances that a family with 4 children will have a boy, a girl, a boy, a
girl; in that order.
(c)
What
are the chances that a family with 4 children will have at least 3 boys?
6.
By
backcrossing the following dihybrids, it is shown that each produces
recombinant gametes in the percentages indicated:
AaBb
→ 31% XxBb →
5% AaYy → 14%
XxYy
→ 22% AaXx → 36%
From this data determine the linear sequence and relative spacing of
each gene locus on the chromosome and draw the proper chromosome map.
a.
Families
were founded by six men from the Bounty, of whom three were blue‑eyed,
two brown‑eyed (but heterozygous for blue) and one homozygous brown‑eyed;
and that there were two Tahitian men and eight women, all homozygous for brown.
b.
The
blue‑eyed, brown alternative depends on a single autosomal allelic pair
with brown (B) dominant and no modifiers of eye color capable of obscuring this
alternative were segregated in this population.
c.
Intermarriage
among the residents of Pitcairn Island was at random with respect to eye color,
with approximately equal numbers of descendant surviving from each type of
marriage.
7.
Gene
frequencies in this population:
(a)
What
were the gene frequencies for the eye color alleles among the six white men,
two Tahitian men, and eight Tahitian women; who, we have assumed, established
families on Pitcairn Island?
(b)
Was
this population at equilibrium in this regard at the time of its establishment?
(c)
What
genotypic and phenotypic proportions would you expect to prevail in this
population after it had reached equilibrium?
8.
The
following table illustrates the predicted results of the first generation of
marriages on Pitcairn Island. Fill in the blank spaces (indicated by question
marks <?>) on the table, then answer the questions below.
Type of
marriage Number of Number of progeny*
female
X male marriages BB Bb bb
BB X BB 3 6 0 0
BB X
Bb 2 2 2 0
BB X
bb <?> <?> <?> <?>
‑‑‑‑‑ ‑‑‑‑‑ ‑‑‑‑‑ ‑‑‑‑‑
TOTALS 8 8 <?> <?>
*Assuming two children per marriage.
(a)
Is
the population at equilibrium after one generation? Explain.
(b)
Is
this inconsistent with the statement that equilibrium is achieved in one
generation of random matings? Why?
9.
In
fowl, genotype rrpp gives single‑comb; R–P– gives walnut‑comb; rrP–
gives pea‑comb; and R–pp gives rose‑comb.
(a)
What
comb types will appear, and in what proportions, in F1 and F2, if single‑combed
birds are crossed with birds of a true‑breeding walnut‑combed
strain?
(b)
What
are the genotypes of the parents in a walnut X rose mating from which the
progeny are 3 rose: 3 walnut: 1 pea: 1 single?
In
man, Hemophilia is the "bleeders disease", a disease in which the
time required for blood to clot is greatly prolonged; depends on the recessive
allele of a sex‑linked gene. There are at present about 40,000 cases of
the disease in the United States. In the following questions, let h+ be the allele for normal
clotting time; h‑ the allele for hemophilia. (Note: In all problems involving sex linkage, list the phenotypes of
sons and daughters separately.)
10.
A
man whose father was hemophilic, but whose own blood‑clotting time is
normal, marries a normal woman with no record of hemophilia in her ancestry.
What is the chance of hemophilia in their children?
11.
A
woman whose father was hemophilic, but who is not herself a
"bleeder", marries a normal man.
(a)
What
is the chance of hemophilia in their children?
(b)
If
these parents have normal son, what is the probability that their next son will
also be normal?
(c)
If
this family has two sons, what is the probability that they will both be
normal?
(d)
If
this family has six sons, what is the probability that three will be normal and
three hemophilic?
(e)
This
family is expecting a child. What is the probability that it will be normal?
That it will be hemophilic?
12.
What
is the chance of hemophilia among the sons of a daughter of the marriage in
question 11, if she marries a normal man?
13.
Why,
in all of these hemophilia problems, were there no hemophilic females
considered?
In
mice, there is a set of multiple alleles of the gene for albinism. Four of
these alleles, listed in order of decreasing amount of color in the hair of
homozygotes, are: C = full color (wild type); cch = chinchilla; cd
= extreme
dilution; and c = albino. This gene is not sex‑linked and you may assume
that each allele is dominant to those below it in the list.
14.
Diagram
a cross between a wild‑type mouse heterozygous for extreme dilution and a
chinchilla mouse homozygous for albinism.
15.
Another
nonsex‑linked gene affecting hair color in mice has alleles B– for black
hair (wild type) and bb for brown hair (recessive type). Expand your diagram
for the preceding question to include the information that both of the mice
crossed are heterozygous for brown (Bb).
16.
Suppose
that after the crosses indicated in the following Drosophilia; the following
were obtained in F2:
wild 292
curled 9
curled, spineless 92
spineless 7
-------
TOTAL 400
Curled wings (cu) and spineless bristles (ss) are autosomal characters in Drosophilia. The genes giving rise to these characters are both in chromosome #3.
(a)
Starting
with a wild‑type female and a curled, spineless male; prepare a diagram
showing parents and progenies through an F2 generation.
(b)
Remembering
again the special circumstance of no crossing‑over in male Drosophilia,
is there some reasonably straight‑forward way to arrive at an estimate of
the map distance between cu and ss?
Explain your method and calculate the distance.
17.
In
cats, black is due to a pair of genes BB. The recessive is bb; a yellow cat.
The coat color, tortoise‑shell is Bb. This is a sex-linked trait in cats.
(a)
A
female yellow cat is bred to a black male. What will be the ratio of kittens?
(b)
What
is the expected ratio of a cross between a tortoise-shell female and a black
male?
(c)
Occasionally
a sterile tortoise‑shell male cat is found. Explain.
18.
Purebred
Holstein‑Friesian cattle are black and white. A recessive allele that,
when homozygous, results in red and white is present but rare in this breed.
Red and white calves are barred from registration, and therefore it is
economically important to avoid using for breeding purposes black and white
individuals that carry the undesirable recessive allele hidden in the
heterozygous condition. How might you detect such heterozygosity in a bull to
be used extensively in artificial insemination?
19.
For
some reason (not yet fully understood) more male babies are conceived than
female. By childhood, the number of boys and girls is the same. Can you think
of a possible explanation for the higher mortality (both before birth and
after) of male babies?
Thalassemia is a type of human anemia
rather common in Mediterranean populations, but relatively rare in other
peoples. The disease occurs in two form, called minor and major, the latter is
much more severe. Severely affected individuals are homozygous for an aberrant
gene; mildly affected persons are heterozygous. Persons free of this disease
are homozygous for the normal allele.
20.
A
man with Thalassemia minor marries a normal woman. With respect to Thalassemia,
what types of children and in what proportions, may they expect?
21.
Both
father and mother in a particular family have Thalassemia minor. What is the
chance that their baby will be:
(a) severely affected?
(b) mildly affected?
(c) normal?
22.
An
infant has Thalassemia major.
(a)
From
this information, what possibilities might you expect to find if you checked the infant’s parents for
anemia?
(b)
Thalassemia
major is usually fatal in childhood. How does this fact modify your answer to
the question above?
23.
The
following questions refer to general genetic situations, in which the
particular phenotypes are not specified.
(a)
A
particular cross gives in F2 a modified dihybrid ration of 9:7. What
phenotypic ratio would your expect in a testcross of the F1’s?
(b)
What
phenotypic ratio would you expect from the testcross of an F1 giving
a 13:3 ratio in F2 ?
(c)
A
9:3:3:1 ratio?
(d)
Indicate
the basis of each of the two dihybrid ratios listed in questions (a), (b) and
(c) above.
24.
In
cattle, RR individuals have a red coat; rr a white coat; and the heterozygous
are roan. The polled characteristic is determined by a dominant gene (P), the
horned alternative by genotype (pp).
(a)
What
will be the phenotype of F1 individuals after mating of a white,
horned animal with a red that is homozygous for polled? What phenotypes will be
found in F2, and in what proportions?
(b)
If
the F1 individuals in the question above (a) are backcrossed with
white, horned animals; what phenotypes will be produced and in what
proportions?
25.
What
kind of gametes does an individual of genotype Ww produce?
26.
What
gametes does an individual of genotype WwXx produce?
27.
In
four-o’clocks red is incompletely dominant to white, and bush is dominant to
lazy (a creeping vine characteristic). What is the genotype ratio of an F1
X F1 cross?
28.
What
is the genotypic ratio of a cross between AabbCcDd X AabbccDD?
29.
What
is the genotypic ratio of a cross between TtSsRr X ttssrr? What is such a cross called?
30.
What
is the genotype of the progeny from the following cross: AaBbCc X AaBbCc?
31.
Assume
that the frequencies of the blood group genes for O, A, and B are 0.65, 0.25,
and 0.10 respectively in computing the frequency of individuals with the
various types of blood.
(a)
Calculate
the frequency of the A, B, AB, and O blood types in the human population.
(b)
Henry
is blood type O, and so is his mother. What blood types may his father have?
(c)
What
are the possible blood types of children resulting from the marriage of a type
O father with a type AB mother?
32.
If
the average proportion of heterozygotes in a population is ¼, what would it be
after 3 generations of inbreeding?
33.
Give
the gene content of the different eggs produced by a woman whose genotype is
JjKkLl.
34.
In
the fruit fly, sepia-eye is recessive to red-eye, and curved-wing is recessive
to straight-wing.
(a)
If
a true breeding sepia-eyed, straight-winged fly is mated with a true breeding
red-eyed, curved-winged fly; what phenotypes will appear in the F1
generation?
(b)
If
the two F1 flies are allowed to mate, what phenotypes will occur in
the F2 generation and in what ratio?
(c)
How
many different genotypes will occur in the F2 generation described
in question (a) above?
35.
When
plant A (a pure breeding tall) is crossed with plant B (a pure breeding short),
the offspring are all intermediate in height. When two of the offspring are
crossed, 1/16th of the next generation is as tall as the tall
grandparent. How do you account for this phenomenon?
36.
Neglecting
crossing-over, how many different kinds of ascospores can Neurospora crassa form by random assortment of its chromosomes, if
seven is the haploid number?
37.
What
sequence of bases in messenger RNA will be coded by the following triplets in
the DNA molecule: GCTCATCCAAAAAGT?
38.
It
is possible to buy sorghum seeds ¾ of which germinate into green seedlings and
¼ of which germinate into albino seedlings. The albino seedlings soon die.
(a)
Why?
(b)
If
A = green and a = albino, what is the genotype of the seeds?
39.
Nondisjunction
may lead to zygotes containing XXY.
(a)
What
sex would this produce in a fruit fly?
(b)
In
a human?
In garden peas the following genetic conditions prevail: The gene for tallness is dominant over the gene for dwarfism. Smooth is dominant to wrinkled. Yellow pod color is dominant to green.
40.
Cross
a heterozygous smooth pea with a heterozygous wrinkled pea.
41.
A
cross between a tall pea and a dwarf pea produced 86 tall plants and 81 dwarf
plants.
(a)
What
is the probable genotype of the tall plant?
(b)
What
is the name for this type of cross?
42. What will be the phenotypes of the offspring of:
(a)
homozygous
yellow X green?
(b)
heterozygous
yellow X green?
(c)
heterozygous
yellow X homozygous yellow?
(d) heterozygous yellow X heterozygous yellow?
43.
If
two animals heterozygous for a single pair of genes are mated and have 200
offspring, about how many will have the dominant phenotype?
44.
Assume
brown-eye dominant in humans over blue-eye. Remember that color-blindness is a
sex-linked trait in humans. A blue-eyed woman whose mother was color-blind
marries a brown-eyed man whose mother was blue-eyed. What are the expected
ratios of their children?
45.
The
Rh factor is carried by a pair of genes designated as DD for Rh+ and dd for Rh-. A woman who is Rh- marries a man who is Rh+ and whose parents were both
Rh+.
What is the possible genotypes of their children and their Rh blood types?
46.
Albinism
in humans is inherited as a simple recessive trait. For the following families,
determine the genotypes of the parents and offspring. When two alternative
genotypes are possible, list both.
a.
Two
non-albino (normal) parents have five children, four normal and one albino.
b.
A
normal male and an albino female have six children, all albino.
47.
Determine
the genotypes of the F2 plants given here by analyzing the
phenotypes of the offspring of these crosses.
|
F2 Plants |
Offspring |
|
(a)
round, yellow X round, yellow |
¾
round, yellow |
|
|
¼
wrinkled, yellow |
|
|
|
|
(b)
wrinkled, yellow X round, yellow |
6/16
wrinkled, yellow |
|
|
2/16
wrinkled, green |
|
|
6/16
round, yellow |
|
|
2/16
round, green |
Is either of the crosses above (a or b) a test
cross?
48.
In
Drosophila melanogaster, gray body color is dominant to ebony body color, while long wings are dominant to vestigial wings. Work the following
crosses through the F2 generation and determine the genotypic and
phenotypic ratios for each generation. Assume the P1 individuals are
homozygous.
a.
gray,
long X ebony, vestigial
b.
gray,
vestigial X ebony, long
c.
gray,
long X gray, vestigial
49.
In
one of Mendel’s dihybrid crosses, he observed 315 smooth, yellow; 108 smooth,
green; 101 wrinkled, yellow; and 32 wrinkled, green F2 plants.
Analyze these data using the chi-square test to see if:
a.
they
fit a 9:3:3:1 ratio
b.
the
smooth:wrinkled traits fit a 3:1 ratio
c.
the
yellow:green traits fit a 3:1 ratio
50.
A
geneticist, in assessing data that fell into two phenotypic classes, observed
values of 250:150. She decided to perform chi-square analysis using two
different null hypothesis: (a) the data fit a 3:1 ratio; and (b) the data fit a
1:1 ratio.
a.
Calculate
the chi-square values for each hypothesis.
b.
What
can be concluded about each hypothesis?
SOURCES FOR PROBLEMS:
Elementary Genetics, Singleton Genetics,
Winchester
Essential Genetics, Klug & Cummings Molecular
Genetics, Horowitz
Human Genetics, Lewis Population
Genetics, C.C. Li
General Genetics, Srb & Owen