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Problem Set 1B
Due Sept 14
1. Solve the problem below with the binomial expansion method. First, indicate the
appropriate bionomial and its expansion. Then use it to answer the following
questions.
(p+q)n = p6 + 6p5q + 15p4q2 + 20 p3q3 + 15p2q4 + 6pq5 + q6
A man and woman marry who are both Rh+ and both heterozygous for the Rh
factor. (Rh+ is dominant over Rh–.) They plan to have 6 children and want to
know the probability that exactly 2 of them will be Rh+ - i.e. that they will have 2
Rh+ and 4 Rh– children. With respect to the binomial expansion above
Define and give the value of p:
the probability of an Rh+ child = 3/4
Define and give the value of q:
the probability of an Rh– child = 1/4
Define and give the value of n:
the number of children in the family = 6
In your binomial expansion above, underline the term would you use to calculate
the probability of 2 Rh+ and 4 Rh– children? Use that term to calculate the
probability below:
15p2q4 = 15[(3/4)2*(1/4)4] ≈ 0.033
2. Red-green color blindness is X-linked recessive. MN blood type is determined by
an autosomal gene with two codominant alleles, LM and LN. A woman whose
father was color-blind has type MN blood. She is going to have a child with a
man who has normal color vision and type MN blood. What is the probability the
child will be a boy, with normal color vision and type M blood?
1/16
3. In a germ-line cell from a human male that is dividing, when do the X and Y
chromosomes segregate?
meiosis I, anaphase
4. A eukaryotic diploid cell from an organism with the ZZ-ZW sex determination
system has two pairs of autosomes and a pair of sex chromosomes, Z and W,
shown below.
A
-a
B --
-b
The individual from which this cell came is (male or female).
female.
What is the probability of a gamete from this individual that has the following
genotype: alleles A and b, chromosome Z?
1/8
5. What is a Barr body?
An inactivated X chromosome, visible in the nucleus of a cell from a female mammal.
6. State and explain the genders of human and Drosophila XXY individuals.
XXY human is male. The SRY on the Y chromosome determines maleness, in
spite of the two X chromosomes. XXY Drosophila is female. Drosophila is
diploid, so there are two of each autosome. If there are two X chromosomes, the
X:A ratio is 1.0, which is a female.
7. You are trying to develop a new species of newt as an experimental model
system. You know that in other species of newt green (G) is dominant to brown
(g) skin color and is determined by a sex-linked gene. You cross brown males to
green females and see that in the F1 all the males are green and all the females
are brown. Which is the heterogametic sex in your species of newt?
Because the F1 females have the recessive brown phenotype they must be
hemizygous (i.e., they inherited a brown allele from their father and no allele from
their mother) therefore the females of this species are the heterogametic sex. We
use the ZZ-ZW nomenclature for species with heterogametic females. Therefore
your F1 females and males are ZgW and ZGZg, respectively.
8. While doing summer field work on a remote Indonesian island you discover a
new genus of lizard closely related to komodo dragons. You attempt to discover
what sex determination system it uses by performing a series of controlled
crosses on the island, using an isolated pair of lizards. Initially, all of your
crosses yield only males (in significant numbers). As fall begins and you prepare
to leave the island you find that your last cross yielded only females (in significant
numbers). Suggest a mode of sex determination that explains this data.
The crosses yielded all males or all females from the same parents. Male and
female progeny were correlated with climatic conditions (summer versus fall).
Environmental sex determination that is dependent on temperature is a likely
explanation.
9. You have discovered a new orange body color gene on the X chromosome of
Drosophila similar to the gene that confers orange coat color in cats. You know
that X chromosome inactivation in female cats heterozygous for the coat color
gene can result in tortoiseshell colored coats. Do you expect to see similar
tortoiseshell body color in Drosophila? Explain your answer?
No, mammals compensate for the dosage difference of genes carried on the X
chromosome between males and females by inactivating one of the X
chromosomes in females. Because the coat color gene in cats is carried on the
X, heterozygous females are chimeric for coat color. In contrast, dosage
compensation is achieved in Drosophila by doubling the activity of genes
expressed on the X chromosome of males. Because there is no X inactivation in
Drosophila one would not expect to see a chimeric expression pattern.
10. Match numbers with the most appropriate letter choice
Pleiotropism (g)
Complementation test (h)
Phenocopy (a)
Incomplete dominance (f)
Cytoplasmic inheritance (d)
Conditional mutation (b)
Sex influenced trait (i)
Genetic maternal effect (j)
Co-dominant (e)
Multi-factorial trait (c)
a. Environmentally produced phenotype
b. Temperature-sensitive
c. Polygenes and environment
d. Non-nuclear
e. Intermediate phenotype
f. Discreet and equal
g. Unrelated effects
h. Homozygous mutations
i. Variable expressivity
j. Maternal genotype
11. In rabbits, an allelic series helps to determine coat color: C (full color), cch
(chinchilla; gray color), ch (Himalayan; white with black extremities), and c
(albino; all white). The C allele is dominant to all others, cch is dominant to
ch and c, ch is dominant to c, and c is recessive to all the other alleles.
This dominance hierarchy can be summarized as C>cch>ch>c. Give the
genotypic and phenotypic ratios expected if rabbits with the following
genotypes are crossed:
a. Ccch x Cch
1 CC (full color); 2 Ccch (full color); 1 cch cch (1 chinchilla)
Ratio 3 full color:1 chinchilla
b. Cch x chc
1 Cch (full color); 1 Cc (full color); 1 ch ch (Himalayan); 1 ch c
(Himalayan)
Ratio 1 full color: 1 Himalayan
c. cchch x chc
1 cch ch (chinchilla); 1 cch c (chinchilla); 1 ch ch (Himalayan); 1 ch c
(Himalayan)
Ratio 1 chinchilla:1 Himalayan
d. Cc x chc
1 Cch (full color); 1 Cc (full color); 1 cch (Himalayan); 1 cc (albino)
Ratio 2 full color: 1 Himalayan; 1 albino
12. A yeast geneticist isolates two different haploid mutant yeast strains,
Strain A and Strain B, which cannot grow unless the amino acid leucine is
added to the growth media. Wild type yeast strains can make their own
leucine, so do not require that it be added to the growth media. She
discovers that each mutant yeast strain contains a single recessive
mutation that leads to the observed leucine-requiring phenotype. When she
crosses the two mutant strains together, she observes that the resulting
diploid can grow without leucine added to the growth media.
Explain the allelic relationship between the mutations in these two strains.
The mutations in strains A and B are NOT allelic, due to the complementation
result observed. Strain A contains a mutation at gene A, which is recessive (a)
and strain B contains a mutation at a separate genetic locus, gene B, which is
also recessive. Strain A contains a wild type B gene and strain B contains a
wildtype A gene. These wild type genes complement the corresponding mutant
alleles in the diploid.
Use the following pedigree for questions 13-17.
I
2
1
3
4
II
1
2
3
4
5
III
1
13. Could the characteristic followed in the pedigree be caused by an autosomal
dominant disease? Why or why not?
No, the offspring of I-1 and I-2 contradict an autosomal dominant inheritance.
14. If the pedigree is for an autosomal recessive characteristic, which individuals are
definitely heterozygous?
I-1, I-2, II-4, II-5, III-1
15. Could the characteristics followed in the pedigree be caused by an X-linked
recessive allele?
Yes, all individuals fit the X-linked recessive inheritance pattern.
16. If the characteristic followed in the pedigree is autosomal recessive, what is III1’s genotype?
definitely heterozygous
17. If the characteristic followed in the pedigree is X-linked recessive allele, what is
III-1’s genotype?
hemizygous for a dominant allele
18. Extensive pedigree analysis on a characteristic shows that:
• the characteristic affects males and females equally.
• two unaffected parents can have an affected child.
• in families in which the parents are unaffected but the children are, 1/4 of the
children are affected.
autosomal recessive
19. Extensive pedigree analysis on a characteristic shows that:
•
•
•
if a woman with an affected father has children with an unaffected man, half of
the sons and none of the daughters are affected.
affected females always have an affected father and an affected maternal
grandfather.
the trait is never passed from father to son.
X-linked recessive
20. Extensive pedigree analysis on a characteristic shows that:
• only males are affected.
• affected fathers always pass the trait to sons.
Y-linked
21. Extensive pedigree analysis on a characteristic shows that:
• males and females are affected equally.
• affected fathers may have affected daughters, but never affected sons.
• half the children of affected mothers are affected.
X-linked dominant
22. Below each person in the pedigree, write his or her genotype, or possible
genotypes, using A for the normal CF allele and a for the disease-causing
recessive allele.
I
Aa
Aa
aa
Tony
AA or
II
Aa
Tina
AA
or
Aa
III
What is the probability that Tony is heterozygous for the CF gene? Explain your
answer.
2/3
Tony’s parents are Aa and Aa. Their children’s genotypes, with probabilities, are:
1/4 AA; 1/4 Aa; 1/4 aA; 1/4 aa. Tony does not have CF, so he cannot be genotype
aa. Of the remaining genotypes, 2/3 are heterozygous and 1/3 are homozygous
dominant.
If the frequency of heterozygotes in the general population is 1/50, what is the
probability that Tony and Tina’s child will have CF? Explain each factor in your
calculation.
2/3 × 1/50 × 1/4 = 2/600, or 1/300
(Tony: heterozygous) × (Tina: heterozygous) × (homozygous recessive from two
heterozygotes)
23. The affected individuals in the following pedigree suffer from a rare bone cancer.
Suggest a mode of inheritance.
Because the trait is rare and occurs in both adopted and non-adopted individuals,
this pedigree is consistent with a trait with strong environmental basis.
24. A DNA sample is tested for base composition. Which choice is one of the
expected results?
a. A = G
b. A + T = C + G
*c. A + C = T + G
d. 50% purines, 50% pyrimidines
e. 25% each base
25. Which of these sequences could form a hairpin?
*a. 5' GGGGTTTTCCCC 3'
b. 5' AAAAAAAAAAAA 3'
c. 5' ACACACACACAC 3'
d. 5' TTTTTTCCCCCC 3'
26. Which of these sequences, if paired with its complementary strand, would be a
palindrome?
a. 5' CCCCCC 3'
*b. 5' CCCGGG 3'
c. 5' CTGCTG 3'
d. 5' TCCCCT 3'
27. If a DNA molecule is 30% cytosine (C), what is the percent guanine (G)?
30%
Refer to the following figure for questions 28-30.
a.
b.
c.
d.
28. Which circle shows a noncovalent bond?
circle b
29. Which circle shows a phosphodiester bond?
circle a
30. Which circle shows a bond that would also be found in an RNA transcribed from
one strand of this DNA?
circle a
Use the DNA sequence shown below for questions 31-33.
3' TACCGTGCGTGACATTAAGCC 5'
31. What would be the sequence of a DNA using the DNA sequence shown as a
template? Write the sequence from 5' to 3', left to right.
5' ATGGCACGCACTGTAATTCGG 3'
32. What would be the sequence of an RNA using the DNA sequence shown as a
template? Write the sequence from 5' to 3', left to right.
5' AUGGCACGCACUGUAAUUCGG 3'
33. Look at your answer to question 43. Some viruses contain an enzyme that
replicates RNA from RNA. What would be the sequence of an RNA using the
RNA sequence as a template? Write the sequence from 5' to 3', left to right.
5' CCGAAUUACAGUGCGUGCCAU 3'
34. While investigating a gene that might be responsible for pathogen resistance in
the plant Arabidopsis, you discover that many of the nucleotides in the gene
sequence are methylated. Which nucleotide (A, T, C or G) is most likely to be
methylated? Draw the structure of this methylated nucleotide. What might this
methylation do to the function of this gene?
(1) Higher eukaryotes, including plants, are typically methylated at cytosine.
(2)
(3) Methylation is often associated with low levels of gene expression.