Download Chapter 4 Extensions of Mendel

CHAPTER 4 EXTENSIONS OF MENDELIAN GENETICS
 Problem 1
In the parakeet, two autosomal genes (that are located on different chromosomes)
control the production of feather pigment. Gene B controls the production of a blue
pigment and gene Y controls the production of a yellow pigment. Recessive
mutations in each gene are known that result in the loss of synthesis of the
respective pigment. Two green parakeets are mated and produce green, blue,
yellow and albino progeny.
a. Based on the information give above, explain this pattern of inheritance. Be sure
to include in your answer:
i. the genotypes of the green parents
ii. the genotypes of all four progeny classes
iii the fraction of total progeny that each phenotypic class represents
b. The parental (green) parakeets were the progeny of a cross between two truebreeding strains. What two types of crosses between true-breeding strains could
have produced the green parents? Indicate the genotypes and phenotypes for each
cross.
.
 Problem 2
Variable expressivity and incomplete penetrance may result from developmental
noise as well as from variations in genetic background and environment.
Developmental noise refers to random variation in the growth and development of
cells and tissues during the generation of the organism. What is a good example of
developmental noise that you learned in the context of X-linked mosaicism?
 Problem 3
“Individuals of identical phenotype may have different genotypes, and vise versa”
Is this statement true or false? Explain using the appropriate genetic terminology.
 Problem 4
In Siamese cats, the albino mutation is responsible for the beautiful coloration of
these animals. This mutation, though, also plays havoc with the neurology of the
Siamese visual system: the development of the nerves linking retina and brain are
affected. Based on this informaion, the albino mutation is:
a. pleiotropic b. epistatic c. incompletely penetrant d. recessive lethal
e. incompletely dominant
 Problem 5
Two genes are involved in determining the color of the foxglove flower. The
dominant allele of the M gene produces a light magenta pigment. The dominant
allele of the D gene causes a darkening of the pigment produced by the M gene,
resulting in dark magenta flowers. The D gene is therefore:
4.1
a. epistatic to the M gene b. dominant to the M gene c. shows variable
expressivity d. is a modifier gene e. none of these answers apply to the D gene
 Problem 6
Marfan syndrome is an autosomal dominant disease state that affects about 1 in
10,000 Americans. Symptoms include myopia, chest deformities, curvature of the
spine, abnormalities of the aorta and a tall, bony stature. All heterozygous
individuals show some symptoms, but most don't show all of the symptoms.
Among the affected members in one family, one brother had poor eyesight and
another 20-20 vision. Both had heart problems. Their mother and sister had poor
eyesight, deformities of the chest, but no heart problems. The terms that best
describe this disease syndrome are:
a. variable expressivity and pleiotropy
b. incomplete penetrance and pleiotropy
c. incomplete penetrance and variable expressivity
d. variable expressivity and multiple allelism
e. variable expressivity and codominance
 Problem 7
A fish from a true-breeding line with high salt tolerance (SS) is crossed to a fish
from a true-breeding line with low salt tolerance (ss). All of the F1 progeny have
medium salt tolerance. A monohybrid cross is performed (F1 x F1). The
following progeny are obtained in the F2 generation:
Results:
202 F2 progeny with high salt tolerance,
403 F2 progeny with medium salt tolerance,
201 F2 progeny with low salt tolerance.
Is this a case of complete dominance or incomplete dominance? Explain.
4.2
 Problem 8
Suppose that salmon has a "temperature-tolerance" gene with two alleles: the coldtolerant allele and the cold-intolerant allele. Also suppose that a native salmon
population is homozygous for the tolerant allele and hatchery salmon are
homozygous for the intolerant allele. A fish and game manager is proposing to
release hatchery fish to interbreed and "supplement" the natural population in
habitat that frequently experiences extreme winter cold snaps.
Using your knowledge of Mendelian genetics, explain why it would be
valuable to know whether the tolerant allele is (1) dominant, (2) incompletely
dominant, or (3) recessive before proceeding. As part of your explanation, predict
the outcome for the F1 generation of a hybrid cross in each of these three possible
cases. (For simplicity, you can pretend that cold-tolerance is the only significant
difference between the wild and hatchery strains).
 Problem 9
Wildtype strains of the haploid fungus Neurospora can make the amino acid
tryptophan (t+ ). A mutant allele (t) renders the fungus incapable of making
tryptophan. Strains carrying a mutant allele (g) of another gene in combination with
either the t+ or the t allele are capable of making tryptophan.
(i) The word that best describes these observations is:
a. codominance b. suppression c. variable expressivity d. pleiotropy
e. none of the above are relevant terms.
(ii) If a genotypically wild-type (t+ g+) strain is crossed with a phenotypically
wildtype but doubly mutant(t g) strain, the fraction of progeny that can synthesize
tryptophan will be:
a. 3/4
b. 1/4
c. 9/16 d. 13/16 e. 1/2
 Problem 10 In a 1945 trial, a woman accused Charlie Chaplin of fathering
her child. The ABO blood types were as follows: woman, A; Chaplin, O; child, B.
When the California jury declared that Chaplin was the father of the child, the
Boston Herald commented: "California has in effect decided that black is white, two
and two are five, and up is down."
a. Do you agree with the jury or with the press? Very briefly explain using
correct genetic terminology.
b. Although we can be sure that the California jury did not know about this, when a
particular genotype (hh) is present at the H gene locus , the genotype of the ABO
bood group is "masked" such that all indivuals appear to be O, no matter what the
genotype at the ABO locus is. (See Russell pg. 282-284)
What single genetic term best describes this phenomenology?
c. Propose a set of genotypes for the woman, Chaplin and the child that would be
genetically consistent with the jury's decision.
d. Present a simple biochemical explanation for why the hh genotype masks the
expression of the A and B antigens and why the h allele recessive to the H allele.
 Problem 11
4.3
Australian coolits have hair color that falls into three categories: dark green, medium
green and white. The results of extensive studies of these coolits are given below:
dark green X dark green ------------->
in some cases: all progeny are dark green
in other cases: 9/16 dark green, 6/16 medium green and 1/16 white
in still other cases: 3/4 dark green and 1/4 medium green
white X white --------> progeny are always white
(i) The best explanation for these data are:
a. Hair color is controlled by one gene with two alleles that show incomplete
dominance.
b. Hair color is controlled by one gene with three alleles that show incomplete
dominance.
c. Hair color is controlled by three genes with two alleles each that show complete
dominance.
d. Hair color is controlled by two genes with two alleles each that show complete
dominance. The genes interact to determine the hair phenotype.
(ii)A number of different true-breeding stocks of medium green coolits are
eventually generated. When two medium greens are crossed from different truebreeding lines are crossed, the phenotypes of the progeny will be:
a. medium green only
b. dark green only
c. both medium and dark green will be produced from any cross
d. the cross could produce either medium green progeny only or dark green
progeny only but not both
e. I don't agree with any of these answers.
 Problem 12
For color to be developed in the aleurone layer of corn, the dominant alleles of four
different genes (A, B, C and R) must be present . In other words, at least one
dominant allele of each of the genes is required for color formation.
If a colored strain of unknown ancestry is tested-crossed to a noncolored line
(homozygous recessive for all four genes) and 1/8 of the progeny are colored, how
many of the genes in the colored strain were heterozygous for the dominant allele?
 Problem 13
Consider two fruit fly genes, on separate autosomal chromsomes, that must
both function normally to get wild-type hair density.
Flies homozygous for the wild-type allele (e) of the Extrahairy gene have
normal hair density. The E allele of this gene is completely dominant to the wildtype allele and results in abnormally hairy flies.
4.4
Flies carrying at least one wild-type allele (B) of the Bald gene produce hair,
whereas flies homozygous for the recessive bald (b) allele are bald. The bald mutant
phenotype is epistatic to the Extrahairy gene.
a. An EEBB female is crossed with eebb male fly and their F1 progeny examined.
The F1 brothers and sisters are then crossed. What are the phenotypes of the
parents and the F1 progeny?
b. For the F2 progeny, give the phenotypic classes produced and their relative
ratios.
c. The following cross is performed with flies that are homozygous for both genes.
A female fly with normal hair density is cross to a bald male. The F1 progeny are
all extrahairy. State the genotypes of the parent flies using the allele designations
indicated above.
 Problem 14
Recall that in bees, fertilized eggs develop into diploid females and unfertilized eggs
develop into monoploid (haploid) males. In bees, the wings of worker females may
be droopy or normal; those of the males (drones) are never droopy. In hives where
droopy-winged workers occur, a number of the unfertilized eggs fail to hatch.
(Within a hive all individuals are the offspring of a single mating between a queen
and a male.) Why do droopy-winged individuals occur among females only? How
can you explain the dead-embryos?
4.5
 Problem 15
In chickens, the "creeper" phenotype results from a mutation that alters
embryological development to produce crooked, shortened legs and wings. A
chicken breeder attempting to get true-breeding line of creepers keeps crossing
creeper males and females. But, whenever two creepers are mated, the progeny
always includes normal non-creeper chicks, creeper chicks and a number of
unhatched eggs. Two possible explanations are:
#1) The chickens are homozygous for the creeper mutation which shows
incomplete penetrance and variable expressivity due to developmental noise
associated with the embryological events in the organism.
#2) All of the chickens are homozygous for the creeper mutation which shows
incomplete penetrance and variable expressivity resulting from a modifier allele in
the genetic background of the parents. The parents of this cross are heterozygous
and the modifier allele shows incomplete dominance in the heterozygotes. The
modifier allele acts more like a suppressor when homozygous.
a. Using proper genetic terminology, propose a third (completely different)
genetic explanations for these observations assuming that the creeper trait shows
complete penetrance.
b. If explanation #1 is correct, will the chicken breeder ever be able to generate a
strain that produces only creeper progeny? Very briefly explain your answer.
c. Answer the question in part b for explanation #2
d. Answer the question in part b for explanation #3
e. Propose a simple set of experiments to distinguish between the various
alternatives. Be as explicit as possible about the predicted outcome of the crosses
and write out genotypes wherever possible. (Define allele symbols)
4.6
 Problem 16
The palomino horse has a golden yellow coat with flaxen mane and tail, the
cremello is almost white, and the chestnut is brown. These horses are identical with
respect to the basic color genotype: all are homozygous for a recessive allele (b)
for brown. The table shows the results obtained upon mating these types in various
combinations.
Parents
cremello
chestnut
chestnut
palomino
palomino
cremello
x
x
x
x
x
x
Offspring
all cremello
all chestnut
all palomino
1 chestnut: 1 cremello: 2 palomino
1 palomino: 1 chestnut
1 cremello: 1 palomino
cremello
chestnut
cremello
palomino
chestnut
palomino
a. Describe the genetic control of coat color as revealed by these results, including
the allelic relationships.
b. Diagram the last three matings.
c. If you raised palominos, how would you set up your breeding program? Why?
d. If a chestnut individual is mated with a cremelllo one what is the chance of
getting: (i) a chestnut animal? (ii) a cremello animal? (iii) a palomino animal?
 Problem 17
In mink (Mustela vison), as in most other mammals, many
genes affect the color of the coat. At least 13 genes are known to determine coat
color in mink. The results of crossing wild-type with platinum (blue-grey) and
aleutian are shown below.
Parents
F1
F2
wild-type x aleutian
wild-type
3 wild-type: 1 aleutian
platinum x aleutian
wild-type
133 wild-type
41 platinum
46 aleutian
17 sapphire
wild-type x platinum
wild-type
3 wild-type: 1 platinum
a. Why is a dihybrid ratio obtained in the third cross and not in either of the first
two?
b. Explain how these results show that there is independent assortment but not
independent gene action.
c. What results would you expect if the wild-types from the cross between
platinum and aleutian were crossed with the sapphires? Why?
 Problem 18 Radishes (Raphanus sativus) may be long, round, or oval in
shape. The color may be red, white, or purple. A long, white variety crossed with
4.7
a round, red one produced oval, purple F1’s. The F2 progeny segregated into nine
phenotypic classes in the following numbers:
long, red
long, purple
oval, red
oval, purple
round, red
long, white
round, purple
round, white
oval, white
9
15
19
32
9
8
16
8
16
a. How many allele pairs are involved? Do they assort independently? What
phenotypes would you expect in crosses between the F1’s and each of the parental
strains?
b. Give the genotypic and phenotypic ratios expected among the progeny of a
cross between: (1) A long, purple and an oval, purple plant.
(2) An oval, purple and a round, white plant.
c. How many true-breeding varieties of radishes could be established?
d. If oval, purple radishes were commercially preferred, state what lines should be
maintained to produce them most profitably and why.
 Problem 19
In Drosophila melanogaster flies may have red (wild-type) or plum eye color and
normal or stubble bristles (less that half normal length). The progeny of a mating
between two plum, stubble flies gave the following results:
plum, stubble
plum, normal
red, stubble
red, normal
162
80
84
42
a. Suggest a hypothesis to account for these results.
b. Outline how you would test your explanation and indicate the results expected in
your experiment.
4.8
 Problem 20
In rats two independently inherited pairs of alleles Aa and Rr interact as follows:
A_R_
A_rr
aaR_
aarr
Grey
Yellow
Black
Cream
The genotypes above are expressed only in the presence of the dominant allele C of
a third gene; the recessive allelec at this locus causes albinism. Four different
homozygous albino lines, each crossed with a true-breeding grey strain, produced
grey F1’s which in turn produced the F2’s shown.
Albino line
1
2
3
4
Grey
174
48
104
292
Yellow
0
0
33
87
Black
65
0
0
88
Cream
0
0
0
32
Albino
80
16
44
171
a. What type of interaction, if any, is involved in each cross?
b. State the probable genotype of each albino line and give reasons for your
answers.
 Problem 21
A woman, whose father is normal, has a rare autosomal dominant disease. This trait
show about 90% penetrance in heterozygotes. If this woman marries a normal
man, what is the probability that their first child will NOT have the disease?
a. 0
b. 0.5 c. 1.0 d. 0.55
e. 0.45
4.9
 Problem 22
The pedigree shown is of streaked hairlessness in Holstein-Friesian cattle, a
condition characterized by an abnormality of the hair coat in which narrow,
irregular hairless streaks, running transversely aroud the trunk, appear.
How can you explain this pattern of inheritance? Try to come up with at least two
different explanations. Which one is most likely? Give reasons for your decision.
4.10
 Problem 23
A geneticist is studying two different genes on two different chromosomes of
a flowering plant. One gene controls “fuzziness” and the other gene controls
flower color. Each gene has two alleles.
Consider the following results, and determine for each gene whether there is
simple dominance or incomplete dominance. Where there is simple dominance,
determine which allele is dominant. Then write the genotypes of the parents for
each cross using appropriate symbols of your own choosing. (for each gene, a
parent may be homozygous or heterozygous for either allele).
#1
#2
#3
#4
#5
#6
fuzzy red x smooth red
fuzzy red x fuzzy white
fuzzy pink x smooth pink
fuzzy pink x fuzzy pink
red smooth x red smooth
fuzzy pink x smooth pink
fuzzy
red
80
0
20
15
0
10
fuzzy
pink
0
60
40
30
0
20
fuzzy
white
0
0
20
15
0
10
smooth
red
0
0
0
5
80
10
smooth
pink
0
20
0
10
0
20
smooth
white
0
0
0
5
0
10
Define your symbols________________________________________________
#1______________________________________
#2______________________________________
#3______________________________________
#4______________________________________
#5______________________________________
#6______________________________________
On the basis of the analysis above, predict the phenotypes and phenotypic ratios
of progeny from the following crosses.
#7
smooth pink x smooth pink________________________________________
#8
smooth pink x smooth red_________________________________________
4.11
 Problem 24 In guinea pigs, black, sepia, cream and albino are all coat colors.
Individual animals showing these colors (not necessarily from pure lines) were
intercrossed; the results are shown in the table below.
Deduce the inheritance of these coat colors, using gene symbols of your own
choosing. Show all parent and progeny types.
Cross
1
2
3
4
5
6
7
8
9
10
______Phenotypes of progeny_______
black
sepia
cream
albino
22
0
0
7
10
9
0
0
0
0
34
11
0
24
11
12
13
0
12
0
19
20
0
0
18
20
0
0
14
8
6
0
0
26
9
0
0
0
15
17
Parental phenotypes
black x black
black x albino
cream x cream
sepia x cream
black x albino
black x cream
black x sepia
black x sepia
sepia x sepia
cream x albino
 Problem 25
In mink, the wild-type is dark in color; two mutant color
patterns called black cross and royal silver have originated in recent times. The
following data was collected:
1) wild-type (true-breeding) x black cross gives 1 black cross : 1 wild-type
2) wild-type (true-breeding) x royal silver gives 1 royal silver: 1 wild-type
3) When black cross mink from the first cross were crossed with royal silver from
the second cross, the kits appeared in the ratio 2 black cross: 1 royal silver: 1 wildtype.
a. These results do not pernit a distinction between (1) multiple allelism and (2)
independent assortment of two allele pairs. Using your own symbols, show how
these results can be explained by each of these two mechanisms.
b. Some of the black cross progeny of the third cross mated with wild-type
produced black cross and wild-type kits only, whereas two other black cross from
the same cross when intermated produced royal silver and black cross offspring
only. Do these results support either of the hypotheses outlined above? Explain.
c. The two black cross offspring that produced royal silver kits when intermated
were also mated to wild-type mink. A small number of offspring were produced,
consisting of royal silver and black cross kits only.
(1) Which hypotheisis do these results support?
(2) What would constitute critical evidence that royal silver and black cross are
determined by different alleles of the same gene?
 Problem 26
4.12
You are a pet store owner who supplies furry critters. You have one truebreeding "wild-type" stock (black fur) and another true-breeding mutant stock (tan
fur).
You have crossed black x tan critters. The F1 individuals all had black fur.
The F2 individuals showed a 3:1 ratio of black:tan. You named the gene "tan", and
you designated the mutant allele as tn and the wild-type (black) allele as tn+ . You
have done reciprocal crosses, and you are confident that the tn gene is not sexlinked.
Another pet store owner sends you a male with a new phenotype, "brown"
fur. The mother of this brown male was tan (from a pure-bred tan line), but it is not
known who the father was. You mate this brown male to two different females:
CROSS I: brown male x tn/tn female:
Progeny: 2 brown males, 2 brown females, 2 tan males, 2 tan females.
You conclude that the male must be heterozygous for the allele that causes the
brown phenotype. Male critters are known to be XY, so the allele that causes the
brown phenotype does not appear to be sex-linked.
CROSS II: The same brown male x black female (pure-bred wild-type)
Progeny: All nine pups have black fur.
a. You hypothesize that the brown phenotype is caused by a new allele of tn, which
you called tnbr. According to your way of thinking, tn+ is dominant to both tnbr
and tn, and tnbr is dominant to tn. (This would be like the black, sepia and cream
alleles of guinea pigs in the problem set). You could symbolize this hypothesis: tn+ >
tnbr > tn. According to your hypothesis, what is the genotype of the brown male?
b. According to your hypothesis, the black pups of CROSS II should have two
possible genotypes (assume the mother is tn+ /tn+ ). What are they?
c. You discuss the above results and your hypothesis with your friend. Your friend
points out a totally different hypothesis that could have accounted for these results
which is that the brown male has a dominant mutant "Suppressor-of-tan" allele (St),
located on a different chromosome from the tan gene. According to your friend, the
brown male mouse could be tn/tn ; St/St+ , and the hypothetical St mutation could be
partially suppressing the tn/tn mutant phenotype. According to this view, the pups
from CROSS II also have two possible genotypes (Assume that the mother in
CROSS II is tn+ /tn+ ; St+ /St+ ). What are they?
4.13
 Problem 26 (continued from previous page):
d. You want to try to distinguish who is on the right track. After further thought,
you decide that you can get some useful information if you cross the black pups
from CROSS II to pure-bred tan individuals, and score the progeny of each pair.
First, you predict what results you would expect according to your hypothesis.
Since there would be two possible genotypes for the black pups, you have to make
separate predictions for each genotype. For each cross, write the predicted
genotypes and phenotypes of progeny, and the expected phenotypic ratio.
e. Next, you make the predictions for what you would get if your friend's
hypothesis is correct. For each cross, write the predicted genotypes and phenotypes
of progeny, and the expected phenotypic ratio. (Ignore sex of progeny)
f. Finally, you obtain results for three of the crosses:
black
Progeny
brown
tan
black pup #1 x tan
4
0
5
black pup #2 x tan
4
2
3
black pup #3 x tan
2
1
0
Are all of these results consistent with your hypothesis?
Are all of these results consistent with your friend's hypothesis?
4.14