Problem Set 1B Answers
9-11-06
1. A geneticist is studying production of an extremely powerful anti-cancer drug that is
produced by a small herb. This drug, called Tamoximillion, is worth millions of
dollars/kilogram. He has identified two loci involved in production of the drug and has
found that when he does crosses of plants heterozygous at each of these loci, the progeny
appear to be of 4 types: ones that contain Tamoximillion, ones that contain a related drug
called Proximillion with lower tumoricidal activity, ones that contain a related drug
called Notamillion with almost no tumoricidal activity, and ones with absolutely no
tumoricidal activity. He has figured out the pathway shown below, partially because he
had determined the ratios at which he obtained each of the plant phenotypes. What is the
phenotype he saw for each genotype listed below? What were the phenotypic ratios he
saw?
Genotype:
A_B_
A_bb
aaB_
aabb
Phenotype:
Prox. present Tamox. present
No tumoricid. act. Notamil.present
Compound with
No antitumor
activity
AProximillion
Tamoximillion
Compound with
No antitumor
activity
bb
Notamillion
Ratios of the different phenotypes:
__3___No tumoricidal activity ___9__Proximillion present
__1___Notamillion present
__3___Tamoximillion present
2. How many loci do all of the ABO blood type alleles map to? ____1 locus______
3. What does a complementation test allow you to determine? (Need only one sentence
or phrase)
Determine whether two mutations are at the same locus or different loci.
4. What is the difference between sex-influenced traits and sex-limited traits? (One or
two sentences)
Sex-influenced traits are encoded on autosomes and preferentially expressed in
one sex over the other. Sex-limited traits are also encoded on autosomes but are
expressed in only one sex.
5. Listed below are blood types for two children and their mothers. Give all possible
genotypes and blood types for the father of each child.
1
Child’s blood type
a. A
b. O
Mother’s blood type
A
B
a. The child is either iAiA or iOiA. The mother contributed an iA allele. The Father
could therefore have contributed either an iA or an iO allele, and thus can have
either type A blood (with iAiA or iOiA genotype) or type O blood (with iOiO
genotype) or type B blood (with iOiB) or type AB blood (with iAiB).
b. Child must be iOiO (because the iO allele is recessive to both iA and iB alleles). The
Mother must be iO iB. The father, who contributed an iO allele, could be either type A
(iOiA genotype), type B (iOiB genotype), or type O (iOiO genotype).
6. Cystic fibrosis is an autosomal recessive disease that includes a build-up of mucous in
the lungs, resulting in problems with breathing and recurrent lung infections. Bob and
Mary do not have cystic fibrosis and have one son and one daughter. Bob’s mother died
of cystic fibrosis shortly after he was born. Bob has a normal sister and a brother with
cystic fibrosis. Mary’s parents are normal, but she has a brother with cystic fibrosis.
(For the purpose of this problem assume that any individual who has the genotype for this
disease is already showing symptoms of the disease.)
A. Considering only the information provided above, draw a pedigree which
includes all the individuals of this family. Use shaded symbols to indicate
affected individuals and a dot in the middle of the symbol to indicate obligate
carriers.
B. Considering only the information above, what is the probability that Bob is a
carrier?
Probability is 1
C. What is the probability that Mary is a carrier?
2
Probability is 2/3. Mary does not have the disease, so she cannot be cc, so
her only possible genotypes are: Cc, cC, or CC (= 2/3 of being Cc and 1/3
of being CC).
D. Considering only the information above, what is the probability that Bob and
Mary’s oldest child has cystic fibrosis?
Probability that Mary is Cc x prob. that when Cc she passes on a c gamete x prob. of Bob passing on a c gamete
2/3
x
½
x
1/2 = 1/6.
This could also be done with modified Punnett squares as we did in lab.
The next 4 statements are FALSE. Cross out the incorrect information in each sentence
and indicate the correction below. You do not need to rewrite the whole sentence, only
the part(s) you strike out. It is not a correct answer to just add a negative statement to the
sentence.
7. A person with Turner syndrome (X,O) is female and shows only Barr body in her
cells.
no
8. Gender in is determined by the ratio of autosomal sets to the number of X
chromosomes.
Drosophila (or fruit flies)
9. When Mendel did dihybrid crosses, one of the characteristics exhibited in the F2
generation that helped him come up with the idea of independent assortment was
phenotypic ratios of .
9:3:3:1
10. When a gene shows imprinting, the imprinted form of the gene comes from the father
and is by methylation of the DNA.
Inactivated
sometimes
11. Dosage compensation in human females is achieved by random inactivation of one
of the X chromosomes and involves coating of the inactive X by ___Xist RNA_______.
This leads to what kind of modification of parts of the DNA? (One word or phrase)
Methylation of the DNA
12. Consider a cross: MmQqJj x MmQqjj.
A. What is the probability of getting an offspring with the genotype
MMQqjj?
¼ x 1/2 x 1/2 = 1/16
B. What is the probability of getting an offspring with the phenotype
mQj?
¼ x ¾ x 1/2 = 3/32
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13. Two pea plants with purple flowers are crossed. Among the offspring, 63 have
purple flowers, and 17 have white flowers. Use a chi-square test to compare the observed
numbers with a 3:1 ratio and determine the probability that the difference between
observed and expected could be a result of chance.
A. What chi-square value did you get?
B. What probability does that correspond to?
C. Does that mean that the hypothesis of a single locus with simple dominance is
still a good hypothesis, or that the hypothesis should be rejected?
Added note: You do not need to memorize the chi-square equation.
There are 80 total offspring, so you would expect (3:1) : 60 of the dominant phenotype
(purple) and 20 of the recessive phenotype (white). Thus, observed minus expected = 6360 for one class and 17-20 for the other class. Plugging the numbers into the chi-square
equation: (63 – 60)2/60 + (17 – 20)2 /20 = 0.15 + 0.45 = 0.6 (the chi-square statistic).
Degrees of freedom = the # of classes (phenotypes) – 1 = 2 – 1 = 1. Looking at
the chi-square table for one degree of freedom, we see that P is only less than 0.1 here
(between 0.1 and 0.5), so the probability (using the standard critical cut-off value of P
needing to be less than 0.05) that this variation is merely due to chance is high enough
that we do not reject the hypothesis. We thus conclude that the observed differences
between observed and expected progeny numbers could be due to chance. So the
hypothesis that both purple parents are heterozygous at the single gene locus controlling
flower color (at least for purple and white) is still a reasonable hypothesis.
14. White eyes in Drosophila are an X-linked recessive trait. Red eyes is the X-linked
dominant form. If a Red eyed, heterozygous female fly is crossed with a red eyed male
fly, what fraction of their female offspring will have white eyes?
What fraction of their male offspring will have white eyes?
XR
Xw
R
Y : YX
YXw
R
R R
X : X X XRXw
Half the males will have white eyes
What fraction of the female offspring will have red eyes AND be a carrier of the white
allele?
Half the females with have red eyes and be a carrier of the white allele.
15. Match each of the following genetic phenomena with the best description of how the
phenotype is determined.
___e____Cytoplasmic inheritance
___f____Genomic imprinting
___b____Sex-influenced characteristic
___a____Genetic maternal effect
___d____Sex-limited characteristic
___c____Sex-linked characteristic
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Phenotype determined by:
a. nuclear genotype of the maternal parent
b. genes more readily expressed in one sex than the other.
c. genes located on sex chromosomes.
d. autosomal genes whose expression occurs only in one sex.
e. genes are normally inherited from the mother.
f. genes whose activity is affected by the sex of the parent from which they came.
16. What kind of traits tend to skip generations?
Autosomal recessive traits and sex-linked recessive traits
17. What kind of traits appear only in males and are passed along only by males?
Y-linked traits
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