Problem Set 2 KEY

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LS4-1 Problem Set 2 KEY
Correct numerical results are not answers to the problems. Assign symbols for genes and
phenotypes for alleles (e.g. “Yellow (Y) gene, Y allele = yellow and y allele = green”) that you
use in your answer. You are encouraged to use a Punnett square, where appropriate, in your
answer. Write out fractions and do math with labels to explain your numerical answer (e.g.
“1/2 boys x 1/2 affected = 1/4 affected boys”).
1. {derived from Problem 3.13} In a population of rabbits, you see four different coat colors:
albino (A), chinchilla (C), himalaya (H) and dark tips (T). To understand the inheritance of
coat colors, you cross individual rabbits with each other and note the results in the following
table.
Cross
1
2
3
4
5
6
7
8
9
10
Parental Phenotypes
H x H
A x H
T x T
C x H
C x C
C x H
A x C
A x A
A x T
C x T
Phenotypes of progeny
1/4 A : 3/4 H
1/2 A : 1/2 H
1/4 C : 1/4 H : 1/2 T
1/2 C : 1/2 T
1/4 A : 3/4 C
1/2 H : 1/2 T
1/2 A : 1/2 C
All A
1/2 C : 1/2 H
1/2 C : 1/2 T
(a) Define gene symbol(s) and allele(s) to explain coat color in this population of rabbits.
C = coat color gene, allelic series: C (chinchilla) = H (himalaya) > a (albino)
C and H are codominant alleles, and the CH heterozygote has dark tips (T).
{This allelic series is similar to inheritance of A, B and i blood type alleles.}
(b) Give genotypes for all individual parents in the ten crosses.
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
Ha x Ha
Ha x aa
CH x CH
Ca x HH
Ca x Ca
CC x Ha
Ca x aa
aa x aa
aa x CH
CC x CH
(c) What kind of progeny would be expected, in what proportions, if you were to cross the
chinchilla offspring in cross #7 and the himalayan offspring in cross #2?
H from #2 is Ha, and C from #7 is Ca.
from cross #7
C
CH
dark tips
Ca
chinchilla
H
from
cross #2
a
a
Ha
himalayan
aa
albino
1/4 dark tips : 1/4 chinchilla : 1/4 himalayan : 1/4 albino
2. {derived from Problem 3.7} See the four pedigrees for familial hypercholesterolemia (FH),
an inherited trait in humans that results in higher than normal serum cholesterol levels
(measured in milligrams of cholesterol per deciliter of blood [mg/dl]), in the textbook problem
3.7. People with serum cholesterol levels that are roughly twice normal have a 25 times
higher frequency of heart attacks than unaffected individuals (yellow symbols). People with
serum cholesterol levels three or more times higher than normal have severely blocked
arteries and almost always die before they reach the age of 20.
(a) In the four Japanese families, what is the most likely mode of inheritance of FH? Either
draw the pedigrees and assign genotypes to all individuals or make a list of all individuals,
labeling individuals by generation (starting with “I” for the oldest) and number from left to
right (1, 2,…) in each generation and assign genotypes.
FH appears to be a dominant trait, vertical inheritance from parent to child.
A = dominant affected allele, a = recessive unaffected allele
The pedigree would also suggest that the two alleles express incomplete dominance.
Family 1
Person
I-1
I-2
I-3
II-1
II-2
II-3
Genotype
aa
Aa
Aa
Aa
Aa
AA
Family 2
Person
I-1
I-2
I-3
I-4
II-1
II-2
II-3
II-4
II-5
II-6
II-7
II-8
II-9
III-1
III-2
III-3
III-4
III-5
Genotype
aa
Aa
aa
aa
Aa
aa
Aa
Aa
Aa
Aa
Aa
Aa
Aa
aa
Aa
AA
Aa
Aa
Family 3
Person
I-1
I-2
II-1
II-2
Genotype
Aa
Aa
AA
AA
Family 4
Person
I-1
I-2
II-1
II-2
II-3
II-4
II-5
II-6
III-1
III-2
III-3
III-4
III-5
III-6
III-7
III-8
III-9
Genotype
aa
aa
Aa
Aa
aa
aa
Aa
Aa
aa
Aa
Aa
AA
aa
aa
aa
AA
aa
(b) Are there any individuals that do not fit your hypothesis? What might account for such
individuals?
Unaffected I-3 and I-4 in family 2 have three affected offspring, and unaffected I-1 and I2 in family 4 have two affected offspring. Possibly the diet of at least two of these
unaffected individuals masked the phenotype of FH in parents of these affected
offspring. Also, these genotypes could, in fact, express overlapping ranges of serum
cholesterol.
(c) Why do individuals in the same phenotypic class (normal or the two affected classes)
show a range of levels of serum cholesterol?
Presumably, serum cholesterol is affected by other factors, such as diet and interaction
with other genes.
3. {derived from Problem 3.21} A pure-bred black mare was crossed to a pure-bred chestnutcolored stallion and produced a chestnut son and a chestnut daughter. The two F1 horses
were mated to each other several times, and they produced a six offspring with three
different coat colors: two bay, two chestnut and two liver.
(a) Could coat color be explained by one gene? by two genes? How? (Assign phenotypes for
genotypes of coat color in the described crosses.)
A cross between pure-bred parents would give one heterozygous genotype in the F1.
Coat color could not be explained by one gene. The parental cross could be explained as
one color (C) gene, where the dominant C give chestnut, and the recessive c give black:
CC x cc ---> Cc. The appearance of two more colors in F2 horses make one coat color gene
impossible because the phenotypes of all three possible genotypes –CC, Cc and cc are
known.
A parental cross with two genes with two alleles, one dominant and one recessive,
can produce as many as four phenotypic classes in the F2. Because there are four colors,
each phenotypic class is expressing a different color. Because the F1 is chestnut, like its
parent, the parental cross must have been AA BB (chestnut) x aa bb (black) ---> Aa Bb
(chestnut). {The alternative parental cross, AA bb (chestnut) x aa BB (black) to make
a dihybrid, would produce an F1 genotype Aa Bb with a third phenotype.}
9 A– B– (chestnut) : 3 A– bb (bay) : 3 aa B– (liver) : 1 aa bb (black)
In the F2 there are just three colors, not all four possible colors. The missing color,
the phenotype (black) for the aa bb genotype, is absent presumably because it represents
the rare doubly recessive class.
(b) Mating a bay-colored F2 horse and a liver-colored F2 horse produced a black foal. What
were the genotypes of the two F2 horses?
The bay F2 horse could be either Aa bb or AA bb, and the liver F2 horse could be either aa
Bb or aa BB. Because the black foal must be aa bb, the gametes from either F2 horse must
have been a b, and the genotypes of the bay and liver F2 horses must have been Aa bb and
aa Bb, respectively.
4. {derived from Problem 3.31} In mice, the mutant Ay allele of the Agouti (A) gene is a
recessive lethal allele but is also dominant (to the normal A allele) for coat color --Ay gives a
yellow coat color, instead of the normal agouti color. Colorless albino (white) mice have a
recessive c allele of the Color (C) gene. c is epistatic to all alleles of the A gene. Mating an
albino mouse to a pure-bred agouti mouse produces a litter of four yellow mice.
(a) What were the genotypes of the parental mice?
albino mouse: AyA cc
agouti mouse: AA CC
(b) What was the chance that all four mice would be yellow mice?
1/2 chance inheriting Ay from albino parent, chance for four mice = (1/2)4 = 1/16
Two yellow F1 mice are mated and produce a litter.
(c) What coats colors are possible and at what frequency would they appear?
This is a sib (or self) dihybrid cross. Frequency of litter composition is conditioned by
the lethality of the homozygous Ay mice.
Ay C
Ay C
Ay c
A C
AyAy CC AyAy Cc AyA CC
(dead)
Ay c
(dead)
yellow
yellow
AyAy Cc AyAy cc AyA Cc
AyA cc
(dead)
A C
A c
A c
AyA Cc
(dead)
AyA CC
AyA Cc
yellow
yellow
AyA Cc
AyA cc
yellow
white
6 yellow : 3 agouti : 3 white
Or,
yellow
white
AA CC
AA Cc
agouti
AA Cc
agouti
agouti
AA cc
white
1/2 yellow, 1/4 agouti, 1/4 white
5. {derived from Problem 3.36} The garden flower Salpiglossis sinuate (“painted tongue”)
comes in many different colors. Several crosses are made between true-breeding parental
strains to produce F1 plants, which are in turn self-fertilized to produce F2 progeny.
Cross
1
2
3
4
5
Parents
red x yellow
lavender x yellow
lavender x red
blue x yellow
blue x lavender
F1 phenotypes
all red
all lavender
all bronze
all blue
all blue
F2 phenotypes
102 red, 33 yellow
149 lavender, 51 yellow
84 bronze, 43 red, 41 lavender
177 blue, 43 red, 15 yellow
244 blue, 59 lavender, 20 yellow
(a) Assign and define symbols to explain the genetic inheritance of flower color in painted
tongues.
Codominant AL allele gives lavender color
Codominant AR allele gives red color
Recessive a allele gives yellow color
heterozygote ALAR = bronze color
Dominant B allele gives blue color
Recessive b allele gives color of A gene
(b) Assign genotypes to the parents, F1 progeny and F2 progeny for all five crosses.
#
1
2
3
4
5
Parent genotypes
ARAR bb, aa bb
ALAL bb, aa bb
ALAL bb, ARAR bb
ARAR BB, aa bb
aa BB, ALAL bb
F1 genotypes
ARa bb
ALa bb
ALAR bb
ARa B–
ALa B–
F2 genotypes
3 AR– bb (red) : 1 aa bb (yellow)
3 AL– bb (lavender) : 1 aa bb (yellow)
2 ALAR bb : 1 ARAR bb : 1 ALAL bb
12 (9 AR– B– & 3 aa B–) : 3 AR– bb : 1 aa bb
12 (9 AL– B– & 3 aa B–) : 3 AL– bb : 1 aa bb
12 : 3 : 1 ratios in crosses 4 & 5 suggest that the F1 is a dihybrid cross in which the
dominant allele B is epistatic to alleles in the A gene.
6. {derived from Problem 3.32} A student whose hobby was fishing pulled a very unusual
carp out of Cayuga Lake: It has no scales on its body. She decided to investigate whether this
strange nude phenotype has a genetic basis. She therefore obtained some inbred carp that
were pure-breeding for the wild-type scale phenotype (body covered with scales in a regular
pattern) and crossed them with her nude fish. To her surprise, the F1 progeny consisted of
wild-type fish and fish with a single linear row of scales on each side in a 1:1 ratio.
(a) Can a single gene with two alleles account for this result? Why or why not?
Pure-bred wild type (aa) x nude (Aa) would give a 1:1 ratio but would have the
phenotypes of the two parents, wild-type scales and no scales, and not a new phenotype, a
linear row of scales. No, a single gene with two alleles would not give a new phenotype.
To follow up on the first cross, the student allowed the linear fish from the F1 generation to
mate with each other. The progeny of this cross consisted of fish with four phenotypes: linear,
wild type, nude and scattered (the latter had a few scales scattered irregularly on the body).
The ratio was 6:3:2:1, respectively.
(b) How many genes appear to be involved in determining these phenotypes?
Segregation of four phenotypic classes suggests that at least two genes are involved in
these fish phenotypes. However, the phenotypic ratio is not the typical 9:3:3:1 for two
genes. This suggests that the F1 linear fish are at least a dihybrid.
In parallel, the student allowed the phenotypically wild-type fish from the F1 generation to
mate with each other and observed, among their progeny, wild-type and scattered carp in a
ratio of 3:1.
(c) How many genes with how many alleles appear to be determine the difference between
wild-type and scattered carp.
One gene with two alleles would give two phenotype in a ratio of 3:1 with wild type as the
dominant allele and scattered carp as the recessive allele.
The student confirmed the conclusion of part (c) by crossing those scattered carp with her
pure-breeding wild-type stock.
(d) Diagram the genotypes and phenotypes of the parental, F1 and F2 generations for this
cross and indicate the ratios observed.
Parents:
AA {wild type} x aa {scattered}
or
F1:
Aa {wild type}
F2:
3 A– {wild type}:1 aa {scattered}
F1 wild type
A
AA
wild type
Aa
wild type
A
F1
wild type
a
a
Aa
wild type
aa
scattered
The student attempted to generate a true-breeding nude stock of fish by inbreeding. However,
she found that this was impossible. Everytime she crossed two nude fish, she found nude and
scattered fish in the progeny in a 2:1 ratio –the scattered fish from these crosses bred true.)
(e) Diagram the phenotypes and genotypes of this gene in a nude x nude cross and explain
the altered Mendelian ratio.
Parents:
F1:
Bb {nude} x Bb {nude}
1 bb {scattered} : 2 Bb {nude} : 1 BB {inviable}
Nude is dominant (B) to scattered (b) but is a recessive lethal allele.
The student now felt she could explain all of her results.
(f) Diagram the genotypes in the linear x linear cross performed by the student [in part (b)].
Show the genotypes of the four phenotypes observed among the progeny and explain the
6:3:2:1 ratio.
P:
aa Bb {original nude} x AA bb {wild type}
F1:
1 Aa bb {wild type} : 1 Aa Bb {linear}
F2:
6 A– Bb {linear} : 3 A– bb {wild type} : 2 aa Bb {nude} : 1 aa bb {scattered}
The three A– BB and one aa BB F2 progeny are inviable, distorting the ratios expected for
the a dihybrid cross.
7. {derived from Problem 2.30} Cutis laxa, a connective tissue disorder in which the skin
hangs in loose folds. A couple, who are expecting a child, have affected individuals in their
family. The soon-to-be mother has an affected mother and unaffected father and sister. The
soon-to-be father has an affected brother and unaffected parents, who happen to be first
cousins.
(a) Draw a pedigree and propose how cutis laxa is inherited.
I
1
2
3
4
II
III
1
2
?
3
4
(b) What is the probability that the soon-to-be mother is a carrier?
Having an affected mother (aa), guarantees that the soon-to-be mother is a carrier.
(c) What is the probability that the soon-to-be father is a carrier?
The father’s parents must both be carriers because his brother is affected. There is a 1/2
chance that he would be heterozygous like his parents, except that we know he is
unaffected. Among unaffected offspring there’s a 2/3 chance that he is also a carrier.
(d) What is the probability that the child will be affected by the disease?
2/3 {father is carrier, Aa} • 1 {mother is a carrier} • 1/4 {offspring of two carriers will
be affected} = 1/6.
(e) What is the probability that the child will be a carrier?
[2/3 {father is carrier, Aa} • 1 {mother is a carrier} • 1/2 {offspring of two carriers will
be carrier}] + [1/3 {father is AA} • 1 {mother is a carrier} • 1/2 {inherit affected a
allele from mother}] = 2/6 + 1/6 = 1/2
8. {derived from Problem 2.35} The common grandfather of two first cousins has hereditary
hemochromatosis, a recessive condition causing an abnormal buildup of iron in the body.
Neither of the cousins has the disease nor do any of their relatives.
(a) Draw a pedigree (as much as you know).
?
(b) If the first cousins mate and have a child, what is the chance that the child will have
hemochromatosis (assuming that the unrelated, unaffected parents of the cousins are not
carriers)?
The parents of the cousins must both be carriers because the grandfather was affected
homozygous for the affected allele (aa). The cousins have a 1/2 chance of inheriting the
affected allele from their parents and being carriers themselves.
1/2 male cousin Aa • 1/2 female cousin Aa • 1/4 chance of child being affected (aa)
= 1/16 = 0.0625
(c) How would your calculation change if you knew that 1 out of every 10 unaffected people
in the population was a carrier for hemochromatosis?
The final answer would not change much.
We know that the parents of the cousins were unaffected carriers, so whether
grandmother was a carrier is irrelevant.
Parent matings to produce the cousins would be either:
AA x Aa ---> 0.9 {not carriers in population} 1/2 {carriers} = 0.45
Aa x Aa ---> 0.1 {carriers in the population} 2/3 {carriers} = 0.0667
Chance that cousin is carrier (Aa) = 0.45 + 0.0667 = 0.5167, instead of 0.5
Chance affected child (aa) = 1/4 • 0.5167 • 0.5167 = 0.0667
(d) How would your calculation change in (c) if the mother and father were unrelated (not
first cousins) and only the mother’s grandfather was affected?
Parent mating to produce affected child would be:
Aa x Aa ---> chance child is affected {aa} =
= 0.1 {freq. of carriers} • 1/2 {father, Aa} • 0.5167 {mother, Aa}
= 0.026
9. {derived from Problem 2.36} People with nail-patella syndrome have poorly developed or
absent kneecaps and nails. Individuals with alkaptonuria have arthritis as well as urine that
darkens when exposed to air. Both nail-patella syndrome and alkaptonuria are rare
phenotypes. In the pedigree (shown in textbook problem 2.36), vertical lines indicate
affected individuals with nail-patella syndrome, while horizontal lines denote affected
individuals with alkaptonuria.
(a) What are the most likely modes of inheritance of nail-patella syndrome and alkaptonuria?
What genotypes can you ascribe to each of the individuals in the pedigree for both of these
phenotypes?
Nail-patella syndrome (N) is a dominant because the trait appears vertically in parent
and child; and, alkaptonuria (a) is recessive because the trait appears horizontally in
the final generation (IV), in children of first cousins.
I-1
I-2
A– nn
A– Nn
II-1
II-2
II-3
II-4
II-5
II-6
AA nn
Aa nn
A– Nn
A– nn
Aa Nn
AA nn
III-1
III-2
III-3
III-4
III-5
III-6
AA nn
A– nn
Aa nn
Aa Nn
A– nn
A– nn
IV-1
IV-2
IV-3
IV-4
IV-5
IV-6
IV-7
A– nn
A– nn
A– Nn
A– nn
aa Nn
aa nn
A– nn
(b) In a mating between IV-2 and IV-5, what is the chance that the child produced would have
both nail-patella syndrome and alkaptonuria? Nail-patella syndrome alone? Alkaptonuria
alone? Neither defect?
Prob{child=aa} = (1){II-2=Aa}•(1/2){III-2=Aa}•(1/2){IV-2=Aa}•
(1){IV-5=aa}•(1/2){child=aa} = 1/8
Probability{child=A–} = Probability{child≠aa}= 1 – (1/8) = 7/8
Probability{child=Nn} = (1) {IV-2=nn}•(1) {IV-5=Nn}•(1/2) {child=Nn} = 1/2
Probability{child=nn} = (1) {IV-2=nn}•(1) {IV-5=Nn}• (1/2) {child=nn} = 1/2
aa Nn: (1/8){aa}•(1/2){Nn} = 1/16
A– Nn: (7/8){A–}•(1/2){Nn} = 7/16
aa nn: (1/8){aa}•(1/2){nn} = 1/16
A– nn: (7/8){A–}•(1/2){nn} = 7/16
10. {derived from Problem 3.30} Two pure-breeding strains of corn produce ears with white
kernels and one pure-breeding strain of corn produces ears with red kernels. These strains
were crossed to each other. The F1 plants were all purple. Red, purple and white kernels
were observed among F2 progeny from the F1 self crosses. The results of crosses between
different strains are tabulated:
Cross
1
2
3
Parent crosses
white-1 x white-2
white-1 x red-1
white-2 x red-1
F1 phenotype
purple
purple
purple
F2 phenotypic ratio
9 purple : 7 white
9 purple : 3 red : 4 white
9 purple : 3 red : 4 white
a. How many genes are involved in determining kernel color in these strains? Define your
symbols and give genotypes for all phenotypes that are shown in the above table.
There are three genes, A, B and C.
A gene is homozygous recessive in white-1:
Dominant A allele gives red/purple, recessive a allele gives white.
B gene is homozygous recessive in white-2:
Dominant B allele gives red/purple, recessive b allele gives white.
C gene is homozygous recessive in red-1:
Dominant C allele gives purple, recessive c allele gives red.
Cross 1
Parents: aa BB {CC} (white-1) x AA bb {CC} (white-2)
F1: Aa Bb {CC}
F2: 9 A– B– (purple) : 3 A– bb (white) : 3 aa B– (white) : 1 aa bb (white)
Cross 2
Parents: aa {BB} CC (white-1) x AA {BB} cc (red-1)
F1: Aa Cc
F2: 9 A– C– (purple) : 3 A– cc (red) : 3 aa C– (white) : 1 aa cc (white)
Cross 2
Parents: {AA} bb CC (white-2) x {AA} BB cc (red-1)
F1: Bb Cc
F2: 9 B– C– (purple) : 3 B– cc (red) : 3 bb C– (white) : 1 bb cc (white)
b. Is there epistasis between alleles in different genes? Define the epistasis.
The recessive b allele is epistatic to C, c. The recessive a allele is epistatic to C, c.
c. How would you interpret your results in terms of genes that are involved in the
biosynthesis of purple pigment in kernels? Create a simple model that explains the
biosynthesis of purple pigment.
A gene product
C gene product
White ---------------------> Red ---------------------> Purple
^
|
| B gene product
|
White
Or,
A gene product
B gene product
C gene product
White ---------------------> White ---------------------> Red ---------------------> Purple
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