Dr. Ballard
BIO 184
Page 1
Fall 2006
EXAM 3
KEY
1. (12 points)
In Vulcans, skin color is under the control of a single autosomal locus: G = green and g = yellow. Ear
shape is under the control of a different autosomal locus: P = peaked and p = round.
You wish to determine if these two genes are on the same chromosome and, if so, how close
together they are.
a.
What cross would you perform in order the map the genes? (Assume that you can set up
controlled crosses in Vulcans.)
Gg,Pp x gg,pp
b.
The cross you set up in part a produces the following offspring:
567
42
534
38
green skin, round ears
green skin, peaked ears
yellow skin, peaked ears
yellow skin, round ears
Is there evidence for linkage? If so, map the loci with respect to one another. Show your
calculation.
Yes, there are two large mirror-image phenotypic classes and two small ones. If there was
no linkage, there would be approximately the same number of offspring in each class.
Map distance = (42 + 38)/(567 + 42 + 534 + 38) x 100 = 6.8 cM
c.
Draw the replicated chromosomes in the dihybrid parent as they would appear just prior to
entering meiosis. Show the locations of all the alleles (G, g, P, and p). Explain how you know
their arrangement.
Gr
gR
gR
Gr
We know this is the arrangement because the two largest phenotypic classes are Gr and gR.
This means that the alleles were carried in these combinations as the dihybrid entered
meiosis, since a dihybrid will always produce more gametes containing the original (parental)
combinations than recombinant combinations.
Dr. Ballard
BIO 184
Page 2
Fall 2006
2. (8 points)
The following three genes have been mapped in Drosophila with the distances indicated below:
eb --------------------------------se----------------sn
4.3 cM
1.3 cM
How many F2 flies (on average) would you need to count in order to see at least one doublecrossover recombinant? Show your calculation.
(0.043)(0.013) = 0.000559
0.000559 (x) = 1 ; x = 1/0.000559 = 1789
So, on average, we would need to count 1789 offspring before we would see our first double
crossover phenotype.
3. (8 points)
Three genes, A F and H, are located on different chromosomes in the round worm, C. elegans. If a
worm with the genotype AA,Ff,Hh was crossed with a worm with genotype Aa,Ff,hh, what fraction
of the offspring would have the phenotype associated with all the dominant alleles?
Use the product rule, and show your calculation.
The phenotype would be A_,F_,H_. (1)(3/4)(1/2) = 3/8
4. (16 points)
Among unicorns, gold horns are highly prized. As a unicorn breeder, you would like to understand
the genetics behind horn color, so you set up a cross between pure-breeding unicorns with gold
horns and pure-breeding unicorns with silver horns. The results of this cross, as well as those of
the F2 generation produced by a cross between the F1s, are given below.
a.
Result 1
F1s
F2s
All have horns that are half gold and half silver
132 gold horns, 261 half gold and half silver, 128 silver horns
How would you interpret these results? How many genes control the trait and why did you
get the results you did?
Horn color is under the control of a single gene with two codominant alleles. H GHG individuals
have gold horns, HGHS individuals have horns that are half gold and half silver, and H SHS
individuals have silver horns.
HGHG x HSHS → F1s (all HGHS)
HGHS x HGHS → F2s ( 1/4 HGHG, 1/2 HGHS, 1/4 HSHS)
Dr. Ballard
BIO 184
b.
Page 3
Fall 2006
Result 2
F1s
F2s
All have silver horns
366 silver horns, 284 gold horns
How would you interpret these results? How many genes control the trait and why did you
get the results you did?
Two genes control the trait. If G is absent (i.e. genotype is gg), then the unicorn has gold
horns, regardless of what the unicorn’s genotype is at the S locus. However, if the unicorn
has “G”, then it will only have silver horns if it also has S. Therefore, a unicorn with the
genotype G_ss will have gold horns.
GGSS (silver) x
ggss (gold) →
GgSs x GgSs → F2s
F2s:
9
3
3
1
G_S_
G_ss
ggS_
ggss
F1s all Gg,Ss (silver)
(silver)
(gold)
(gold)
(gold)
This explains the 9 silver : 7 gold ratio observed among the F2s.
G_
Gold
gg
S_
Gold
Gold
ss
Silver
Gold
5. (10 points)
Caryn’s father, Arthur, has unusually hairy earlobes, a trait that has run through his side of the
family for at least three generations. Caryn’s brother, Nick, also has hairy earlobes, but Caryn is
relieved that she does not. However, Caryn has two sons, both of whom have hairy earlobes.
Provide two different explanations for how this trait could have skipped Caryn only to reappear in
her two sons.
1. The trait is dominant in males and recessive in females due to hormones. Caryn inherited the
mutant allele from her father but it was recessive in her. When she passed it to her sons, it was
dominant in them.
2. Incomplete penetrance. Some environmental factor is at work. Caryn was not exposed to this
factor and so did not get the hairy ears. However, her sons were exposed and therefore have the
phenotype.
Dr. Ballard
BIO 184
Page 4
Fall 2006
6. (16 points)
In horseradish, flower color, seed size, and leaf shape are under the control of three different
genes, all of which are located on the same chromosome.
M = mustard flowers, m = white flowers
L = large, l = small
R = round, r = oblong
A plant heterozygous at all three loci is test crossed with the following results:
1567
1578
13
11
146
139
88
91
a.
Mlr
mLR
MlR
mLr
mlr
MLR
MLr
mlR
Which gene is in the middle? How do you know?
The parentals are Mlr and mLR. The double co’s are MlR and mLr. The gene whose alleles are
switched when comparing these two groups is R/r. Therefore, R/r is in the middle.
b.
Draw a map of the genes, showing the distance between adjacent loci. Show how you
calculated these distances.
8.5 cM
5.6 cM
M/m--------------------------------------R/r------------------------L/l
Distance M-R: P = Mr, mR; R = MR, mr. Map distance = (13+11+146+139)/3633 x100 = 8.5 cM
Distance R-L: P = rl, RL; R = rL, Rl. Map distance = (13+11+88+91)/3633 x 100 = 5.6 cM
7.
(10 points)
Boys and girls with gigantism almost always inherit a deletion of a small portion of chromosome 18
from their mothers. Some of these mothers also suffer from gigantism but others do not. Why are
only some of the mothers affected with the disease? How does this help explain why children with
gigantism always inherit the chromosome deletion from their mothers? Draw a diagram to support
your answer.
Dr. Ballard
BIO 184
Page 5
Fall 2006
This gene is imprinted. It is normally turned off in the father’s germ-line and turned on in the
mother’s. Therefore, if a child inherits the deletion from his/her mother, he/she will have no good
copy of the gene and will have gigantism.
Whether or not a person is affected is therefore entirely dependent on which parent contributed
the mutant chromosome. If a woman inherited the mutant chromosome from her father, she will not
be affected. However, when she passes the mutant chromosome to a child, the child will be.
However, if a woman inherited it from her mother, she will be affected and so will any children who
inherit it from her. This explains why some mothers of children with gigantism are also affected
and some are normal.
This gene is probably a growth factor repressor. The male germ-line turns off the repressor in
order to encourage the embryos to grow large. The females turn on the repressor to encourage the
embryos to remain average in size.
Normal situation: (* = turned off)
18*,18 (Dad)
18*,18 (Mom)
germ-line
All sperm 18*
All eggs 18
If deletion comes from Dad:
18*,18(del) (Dad)
18*,18 (Mom)
germ-line
1/2 sperm 18*
1/2 sperm 18*(del)
All eggs 18
fertilization
All children normal
Dr. Ballard
BIO 184
Page 6
Fall 2006
If deletion comes from a giant Mom:
18*,18 (Dad)
18*,18(del) (Mom)
germ-line
All sperm 18*
1/2 eggs 18
1/2 eggs 18(del)
fertilization
1/2 children normal (18*,18)
1/2 children giants (18*,18(del))
If deletion comes from a normal Mom:
18*,18 (Dad)
18,18*(del) (Mom)
germ-line
All sperm 18*
1/2 eggs 18
1/2 eggs 18(del)
fertilization
1/2 children normal (18*,18)
1/2 children giants (18*,18(del))
Note: To get full points you did not have to diagram all these scenarios, but this is a full
answer in case you were confused by the question.
Dr. Ballard
BIO 184
Page 7
Fall 2006
True (T) or False (F)? (20 points)
If the statement is false, cross out the underlined word and replace it with the correct word.
Example:
T
F
An individual with the genotype aaBbCcDd can produce 8 different type of gametes.
F, Thomas Morgan
T
Forensic Files
Dr. Ballard is hooked on the TV program Baywatch.
Linkage mapping was invented by Gregor Mendel.
Blue blood disease is an example of a dominant disease caused by a poisonous allele.
T
The inheritance of X-linked genes violates Mendel’s First Law.
F;OO
If a man has type B blood and his wife is type O, the children could be BB or BO.
T
Achondroplasia is caused by a mutant human disease allele that is incompletely dominant.
T
Homozygotes for dominant diseases usually arise by inbreeding or assortive mating.
T
When two genes control a trait, Mendel’s Second Law is violated.
F; affected
F; some
T
When a mutation is dominant, heterozygotes are unaffected.
All female tortoise-shell cats have the genotype BB,Oo.
Gene expression can be turned off by methylation of the C residues in the promoter.
T
Mammals with a promiscuous mating system tend to have males with large testicles.
T
Mendel completed his research before Thomas Morgan began his.
F;two
The coat color of Labrador retrievers is controlled by three genes.