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Midterm Answers for Biology 202 (2010)
Dr. Schoen’s section (answers in bold—please see notes below each question for
more detailed explanation, when needed):
(1.) (1 point) “Forward genetics” is a term used to refer to the approach:
(a) In which mutations are induced in an organism’s genome and the investigator
attempts to determine how mutational changes in a particular portion of the
sequence influences the organism’s phenotype.
(b) In which the organism’s external environment is modified and the investigator
attempts to determine how this modification influences the organism’s genotype.
(c) In which crosses are conducted between organisms that have mutant and
wild type phenotypes, and the inheritance patterns of progeny are studied,
eventually leading to the identification of the gene(s) responsible for the
mutant phenotype.
(d) In which transcription of a portion of an organism’s DNA sequence is inhibited
and the investigator attempts to determine how this influences the organism’s
phenotype.
(e) In which the organism’s development is modified and the investigator attempts to
determine how this modification influences the organism’s genotype.
(2.) (1 point) In shorthorn cattle, the genotype R/R causes a red coat, R/r causes a roan
coat, and r/r causes a white coat. A breeder has all colors of cattle and makes the
following crosses:
Cross 1:
Cross 2:
Cross 3:
Cross 4:
Red x Red
Red x Roan
Red x White
Roan x White
What proportions of progeny are expected in the next generation from each cross?
(a) all Red (Cross 1); all Red (Cross 2); ½ Roan and ½ White (Cross 3);
all Roan (Cross 4)
(b) all Red (Cross 1); ½ Red and ½ Roan (Cross 2); all Roan (Cross 3);
½ Roan and ½ White (Cross 4)
(c) all Red (Cross 1); all Red (Cross 2); all Roan (Cross 3);
½ Roan and ½ White (Cross 4)
(d) all Red (Cross 1); ½ Red and ½ Roan (Cross 2); all Roan (Cross 3);
all Roan (Cross 4)
(e) None of the above.
(3.) (1 point) Homologous chromosomes pair, synapse, and separate during:
(a)
(b)
(c)
(d)
(e)
Mitosis
Meiosis II
Meiosis I
Mitosis and Meiosis I
Mitosis and Meiosis II
(4.) (1 points) Consider the pedigree below in which the a allele, responsible for the
expression of a rare trait (filled in symbols), is recessive to the A allele. Which
of the following assignments of genotypes to individuals most likely? (Note: Roman
numeral I refers the parents and II to the progeny. Individuals within generations are
numbered left to right in increasing order).
(a.)
(b.)
(c.)
(d.)
I-1: A/a, I-2: a/a and II-1: A/a, II-2: a/a, II-3: A/a, II-4: a/a, II-5: a/a
I-1: a/a, I-2: A/a and II-1: A/a, II-2: a/a, II-3: A/a, II-4: a/a, II-5: a/a
I-1: A/A, I-2: A/a and II-1: A/a, II-2: A/A II-3: A/a, II-4: A/A, II-5: A/a
I-1: XA/Xa, I-2: Xa/Y and II-1: XA/ Xa, II-2: Xa/Y, II-3: XA/ Xa, II-4: Xa/Y, II-5:
Xa/Xa
(e.) None of the above.
Note: Several students wrote that answers (a) and (d) are equally likely. However this
is not the case. Since the trait is rare in the population, then the underlying mutation
must also be rare. For example, suppose that the mutation has a frequency of 0.01 in
the population. Under scenario in answer (a), the probability of sampling an A/a
individual is (from Hardy-Weinberg theory) 2 x 0.99 x 0.01=0.0198, while the
probability of sampling an a/a individual is 0.012 =0.0001. So the probability of
sampling a family in which one parent is A/a and one is a/a is the product of these two
quantities, or 0.00000198.
Under scenario in answer (d), the probability of sampling an XA/Xa mother is
2 x 0.99 x 0.01=0.0198, and the probability of sampling an Xa/Y father is 0.01. Again
using the product rule, the probability of sampling a family with this set of parents is
0.000198, or 100 times more likely than scenario (a).
This question is challenging and tests your ability to put together population genetic
reasoning with pedigree analysis. The idea is that we are searching for the answer that
best fits what we know about the trait (it is recessive and rare), and using quantitative
reasoning to reach the best solution.
I have accepted answer (a), even though it is not the best answer, because some
students were confused as to ho to minimize the introduction of rare alleles into the
pedigree.
(5.) (1 point) Mitosis and meiosis differ in regard to the presence of:
(a) Chromatids
(b) Mutation
(c) Synapsed homologs
(d) Centromeres
(e) Spindles
(6.) (2 points) Five separate nutritional mutants in Neurospora were independently
isolated. They all require compound F to grow. Intermediate compounds in the
biosynthesis of compound F are known and tested for their ability to support the growth
of each mutant. The results are given in the table below, where (+) indicates growth and
(0) indicates no growth. Assuming a linear pathway, what is the order of the compounds
A through F?
Compounds
Mutants A B C D E F
––––––––––––––––––––––––––––––––––
1
0 0 0 + 0 +
2
0 + 0 + 0 +
3
0 0 0 0 0 +
4
5
0 + + + 0 +
+ + + + 0 +
a) EABCDF
b) ECBDAF
c) ECDBAF
d) EACBDF
e) EBACDF
Note: See lecture on Beadle and Tatum’s experiments with Neurospora.
(7.) (2 points) Below is a pedigree for a family with a recessive sex-linked trait. Suppose
we know that the allele for this trait enters the pedigree in the generation of individual A,
and once more in the generation of individual B. What is the probability that the child of
individuals C and D would exhibit the trait?
(a) 0
(b) 1/4
(c) 1/3
(d) 1/2
(e) 1
Note: Some students correctly noted that the individual to the right of individual D
should be affected in this case, where the trait is said to bne sex-linked, the mother of D
is affected, or, in other words, possesses the genotype Xa/Xa (where a denotes the
mutation). So in this case we have a pedigree with incomplete penetrance; the
individual to the right of D is not affected when it might be expected to be. While this
may seem to some to make for an ambiguous question, it does in fact not alter the way
one would approach the analysis of the pedigree. To illustrate this, note that the male
parent of individual C must be genotype Xa/Y, and thus inherited the Xa through his
mother. We are told that the trait enters the pedigree again in the next generation, and
by inspection of the pedigree, we see this is through the mother of D. Given what we
are told about the trait (it is sex linked and recessive), the mother of D must be Xa/Xa.
With an Xa/Y times Xa/Xa cross, and since the sex of the offspring in question is not
known in advance (this is part of the question!), we can say that ¼ are expected to be
affected. There is no basis for selecting any of the other answers.
(8.) (2 points) The genotype of a newly discovered species of tall peas may be denoted as
T/_ and that of short peas as t/t. The genotype of round peas may be denoted as R/_ and
that of wrinkled peas as r/r. Suppose you do not know whether the two genes assort
independently, and so you conduct a test in which you self pollinate a plant that is doubly
heterozygous (T/t ; R/r). You find the following numbers of progeny phenotypes:
Phenotype
Tall, Round
Tall, Wrinkled
Short, Round
Short, Wrinkled
Numbers
80
40
25
15
You conduct a Chi-Square test to determine whether the hypothesis of independent
assortment is supported by these data. In this test, you decide to assume that all
genotypes have equal viability. You find that the Chi-Square statistic has a value of:
(a)
(b)
(c)
(d)
(e)
0.555
6.417
7.778
250
25
Note: Apply the Chi-square formula to the numbers expected under a 9:3:3:1 ratio, to
obtain 7.778.
(9.) (1 point) The critical value of the Chi Square can be thought of as the value of the
test statistic that would be obtained 5% of the time (or less) if the hypothesis being tested
is true. The critical value of the Chi-Square with 4 degrees of freedom is 9.488. The
critical value of the Chi-Square with 3 degrees of freedom is 7.815. The critical value of
the Chi-Square with 2 degrees of freedom is 5.991. The critical value of the Chi-Square
with 1 degrees of freedom is 3.841. For the study described in the question immediately
above, what can we say about how the data pertain to the hypothesis that the two genes
assort independently?
(a) Data such as these would be expected in at least 5% or more of the
experiments conducted when the two genes assort independently, and so we
conclude that the data are compatible with the hypothesis.
(b) Data such as these would be expected in fewer than 5% of the experiments
conducted when the two genes assort independently, and so we conclude that the
data are NOT compatible with the hypothesis.
(c) Data such as these would be expected in at least 5% or more of the experiments
conducted when the two genes assort independently, and so we conclude that the
data are NOT compatible with the hypothesis.
(d) Data such as these would be expected in fewer than 5% of the experiments
conducted when the two genes assort independently, and so we conclude that the
data are compatible with the hypothesis.
(e) None of the above.
Note: Answer (a) provides the definition and interpretation of a non-significant Chi
Square (for three degrees of freedom—the case we are interested in here).
(10.) (1 point) Peas heterozygous for four independently assorting genes are intercrossed.
Each gene has a dominant and recessive allele. What proportion of the progeny is
expected to show all four recessive traits?
(a)
(b)
(c)
(d)
(e)
1/64
1/128
1/256
1/16
1/32
Note: (1/4)4=1/256
(11.) (1 points) For the experiment described above, how many progeny would need to
be grown to ensure that there is a 95% chance of obtaining a single plant that is
homozygous for all 4 recessive alleles?
(a)
(b)
(c)
(d)
(e)
32
64
128
256
765
Note: The probability of obtaining a single plant that is not homozygous for all 4
recessive alleles is 1-(1/256)=255/266. In a sample of n plants, the probability of
obtaining no plants that are homozygous for all 4 recessive allele = (255/266)n. The
probability of obtaining a single plant (or even more) that are homozygous for all 4
recessive allele = 1 - (255/266)n . Setting 1 - (255/266)n = 0.95, we see that n must be
765.
(12.) (1 point) Recombination during meiosis occurs because of:
(a)
(b)
(c)
(d)
(e)
Events taking place during Meiosis II
Crossing over
Independent assortment
Answers (a) and (b)
Answers (b) and (c)
Note: Recombination occurs by BOTH independent assortment AND crossing over, as
pointed out in class and in the text. Independent assortment leads to 50%
recombinants, whereas, crossing over (of linked genes) leads to <50% recombinants.
(13.) (1 points) A cross is conducted between two pure breeding lines of Drosophila.
One parent is homozygous recessive mutations for the mutations sc, ec, and cv. The
other is homozygous for the normal allele at these three loci. The F1 is testcrossed and
the testcross progeny are scored for their phenotype. Below is a table showing the
results:
Phenotype of F2 offspring
sc ec cv
Number observed
1158
sc+ ec+ cv+
1455
sc ec+ cv+
163
sc+ ec
130
cv
sc ec cv+
192
sc+ ec+ cv
148
sc ec+ cv
1
sc+ ec
1
cv+
3248
TOTAL:
Which of the following statements is correct?
(a) The genes are all on the same chromosome their order is: sc -- ec -- cv
(b) The genes are all on the same chromosome their order is: sc -- cv -- ec
(c) The genes are all on the same chromosome their order is: ec -- sv -- cv
(d) The genes are all on two sets of chromosomes with: ec – sv (on the 1st chromosome,
and cv (on the 2nd chromosome).
(e) The genes are all on two sets of chromosomes with: ec – cv (on the 1st chromosome,
and sv (on the 2nd chromosome).
Note: See lecture notes on three-point testcross mapping.
(14.) (1 point) Based on the data from the experiment described above, which of the
following statements is correct?
(a) There were about 6% fewer double crossovers observed than would be expected if
a crossover between one of the gene pairs had no influence on the probability of a
crossover between the adjacent gene pair.
(b) There were about 6% more double crossovers observed than would be expected if
a crossover between one of the gene pairs had no influence on the probability of a
crossover between the adjacent gene pair.
(c) There were about 94% fewer double crossovers observed than would be
expected if a crossover between one of the gene pairs had no influence on the
probability of a crossover between the adjacent gene pair.
(d) There were about 94% more double crossovers observed than would be expected
if a crossover between one of the gene pairs had no influence on the probability of
a crossover between the adjacent gene pair.
(e) There is no way to determine whether the numbers of double crossovers were
more or less frequent than expected given the observed numbers of single
crossovers.
Note: See lecture notes on calculating interference (and interpreting what it means).
(15.) (2 points) Nail patella syndrome is caused by a dominant autosomal mutation. The
ABO blood groups are coded for by an autosomal gene with the alleles A and B being
codominant to each other, and both dominant to allele O. For the pedigree below, what is
the LOD score under the hypothesis that the two loci are linked with recombination
fraction 0.2?
a.)
b.)
c.)
d.)
e.)
1.03089
10.7375
2.8665
0.4573
None of the above
Note: There is one recombinant and 7 parental types.
Take the log base 10 of: [(0.4)7(0.1)]
(0.25)8
(16.) (2 points) Cystic fibrosis is caused by a rare recessive allele and symptoms develop
early in life. Assume that the genotype frequencies in human populations are in HardyWeinberg proportions for the locus in which this mutation occurs. Cystic fibrosis
affects 1 out of every 10,000 individuals. A man who is a known carrier of the cystic
fibrosis mutation marries an unrelated woman who does not have the disease (but could
be a carrier). What is the probability that their first-born child will inherit the disease?
a.) 0.25
b.) 0.00005
c.) 0.00495
d.) 0
e.) 0.5
Note: This question asks you to use population information to determine the
probability that the “unrelated woman” is a carrier, that is, is heterozygous for the
normal and mutant alleles. From the incidence of the disease and the fact that the
genotype frequencies are in HW proportions, we can deduce that the frequency of the
mutation is the square root of 0.0001, or in other words, 0.01. Knowing this, we can
say that the expected frequency of carriers is 2 x 0.99 x 0.01. = 0.0198. From a mating
of two heterozygotes, we expect ¼ to be affected; ¼ x 0.0198 = 0.00495.
(17.) (1 point) Which of the following statements about inbreeding is false?
a.) Complete self-fertilization in a population over five consecutive generations should
reduce the frequency of heterozygotes by a factor of 1/32 compared to the starting
generation.
b.) The expected equilibrium frequency of heterozygotes in a population that has been
inbreeding for many generations, but that otherwise conforms to the assumptions of
Hardy-Weinberg theory, can be determined if one knows the inbreeding coefficient for
the population (and the inbreeding coefficient has remained constant).
c.) Brother-sister mating is expected to alter genotypes frequencies in the population
where it occurs.
d.) The incidence of rare recessive genetic disorders is likely to be higher in a population
with an inbreeding coefficient of 1/16, compared to another, similar population in which
the inbreeding coefficient is 1/32.
e.) A small, isolated population that is kept small for many generations will remain
in Hardy-Weinberg proportions when there is no mutation, migration, or selection.
Note: As pointed out in class and in the text, the conditions in answer (e) lead to
genetic drift, which over time erodes variation and leads to inbreeding. Inbreeding, in
turn, causes loss of heterozygosity with respect to that expected with HW proportions.
(18.) (1 point) Which of the following scenarios is NOT likely to maintain genetic
variation at a locus over many generations of time?
a.) Ongoing mutation of allele A to allele a occurs at rate u=0.0001. Individuals with the
a/a genotype, however, have relative fitness w=0.5 (compared with individuals of
genotypes A/A and A/a, who have relative fitness = 1.0).
b.) Three genotypes B/B, B/b, and b/b, in a large population, have relative fitnesses of 0.8,
1.0, and 0.9, respectively.
c.) Locus A experience heterogeneous selection, such that adults of genotype A/A are
favoured in wet parts of a grassland and adults of genotype a/a are favoured in dry parts
of the same grassland. Progeny of these adults can disperse and re-establish next year in
either wet or dry parts of the grassland.
d.) A large population has a mating system in which first cousins mate.
e.) The population is maintained at a size of 20 individuals, and the mutation rate at
the locus being examined is zero.
Note: Inbreeding in a large population does not lead to loss of genetic variation. It
simply reshuffles it into homozygotes. Alleles are not lost.
(19.) (1 point) If two independently derived recessive mutant alleles that produce similar
recessive phenotypes fail to complement each other, what can you conclude?
a.) The alleles must be on different genes.
b.) They are both mutations of the same allele.
c.) The alleles must be on the same gene.
d.) The alleles are epistatic.
e.) The alleles are both homozygous.
Note: See lecture notes on complementation.
(20.) (1 point) If a single allelic dose of a gene product is needed for a biochemical
pathway to go to completion then:
a.) Epistatic interactions with the null mutation will occur.
b.) Pleiotropy of the null mutation will occur.
c.) Suppression of the null mutation will occur.
d.) The mutation is recessive.
e.) There will be a 9:7 (Wild type:Mutant) inheritance pattern when the
heterozygote is intercrossed with another heterozygote.
Note: The locus is haplosufficient. The mutation is recessive.
(21.) (1 points) Genes A, B, and C are independently assorting and control for enzymes
that are in a biochemical pathway for the production of black pigment. Suppose that A,
B, and C act in the following pathway,
A
B
C
Colourless precursorintermediate 1intermediate 2Black pigment
The alternative (and recessive) alleles of these three genes (a, b, and c) all code for
abnormal products that interrupt the pathway, AND the intermediates 1 and 2 are
colourless.
A black A/A B/B C/C individual is crossed with a colourless individual a/a b/b c/c to
give a black F1. The F1 is self-fertilized. What proportion of the F2 individuals is
colourless?
(a)
(b)
(c)
(d)
(e)
1/64
27/64
37/64
9/64
3/64
Note: For a coloured product to be produced, we must have genotype: A/_ B_ C/_.
All other genotypes are blocked at one (or more) steps, and so the product is colourless.
The probability of genotype A/_ B_ C/_ is: ¾ x ¾ x ¾ = 27/64. But since we are
interested in the probability of a colourless phenotype, not a coloured phenotype, we
want the converse, or in other words: 1-27/64 = 37/64.
(22.) (2 points) You are studying an annual plant population in which selective mortality
occurs after zygote formation, but before the plants mate. Mating is at random with
respect to the genotypes at locus A. The relative fitnesses of the genotypes A/A, A/a, and
a/a are 1, 1, and 1/3, respectively.
At the start of generation t, a large population has allele frequencies p = f (A) = 1/4, and
the population is in Hardy-Weinberg proportions with respect to locus A.
What will the frequency of the genotype A/A be in zygotes of generation t+1?
a) 0.16
b) 0.40
c) 0.10
d) 0.32
e) 0.30
Note: See the lecture on calculating allele frequency change under selection. Some
students missed this question because they calculated the frequency of A/A in
generation t (where it is 0.1), not in generation t+1, as the question asks.
(23.) (1 point) Genetic drift :
(a) occurs when populations are founded by a small number of individuals.
(b) increases homozygosity in populations.
(c) leads to loss of alleles in populations.
(d) Answers (a) and (c) are correct.
(e) Answers (a) and (b) and (c) are correct.
Note: Genetic drift has all thee characteristics.
Dr. Chevrette’s section (answers given below each question):
Question 1 (1 point)
1. Identify the incorrect statement:
a) The DNA sequence flanking microsatellite DNA must be known to be able to
analyze this type of DNA polymorphism.
b) A point mutation occurring in a DNA molecule will be seen as an RFLP only if this
mutation affects the restriction site of a given enzyme.
c) RFLP analysis can be used even if we do not know the sequence of the fragment
to be analyzed.
d) Cesium chloride gradients can be used to isolate telomeric repeated sequences.
e) The length of the VNTRs varies between individuals and even in the same individual.
Answer is d)
Question 2
(2 points)
Please read the following statements carefully:
1)
2)
3)
4)
5)
DNA polymerase pausing and slippage during DNA replication is the main
cause of minisatellite polymorphism.
Microsatellite polymorphisms are generated by misalignment and unequal
crossing-over.
In contrast to eukaryotic DNA, prokaryotic DNA contains many repetitive
elements.
When analyzing DNA with RFLPs, the deletion of a 1 kb fragment in the
middle of a five kb EcoRI fragment detectable with the probe you are using
will always be seen as a 4 kb fragment on the autoradiogram.
Mouse satellite DNAs are located near all telomeres of the mouse
chromosomes.
Now please choose the best answer from below:
a)
b)
c)
d)
e)
Statements 1), 2) and 5) are correct while statements 3) and 4) are false.
All statements are false.
Statements 1), 2) and 4) are correct while statements 3) and 5) are false.
Statements 1), 2), 3) and 4) are false while statement 5) is right.
Statement 4) is right, all others are false.
Answer is b)
Question 3 (1 point)
The following figure represents an ethidium bromide stained gel containing
molecular weight markers (“M”) in the first lane, followed by fragments of DNA from
5 samples (lanes 1 to 5) that have been digested with restriction enzymes and
separated according to their size.
M
1
2
3
4
5
10
kb
8
7kb
6
kb
kb
4
3
kb
kb
1
kb
Which one of the following statements will always be true.
a) This gel could represent the complete digestion of the same DNA with 5 different
restriction enzymes.
b) The 1 kb fragment seen in samples 2 and 3 are identical and will have the same
DNA sequence.
c) If one was to use the 4 kb fragment of sample 1 as a probe in a Southern blot
obtained from this gel, there will be a hybridization signal in samples 1, 3 and 5
only.
d) If one was to use the 7 kb fragment of sample 2 as a probe in a Southern blot
obtained from this gel, there will be a hybridization signal in all samples.
e) If the DNA sequences of samples 1 and 3 are the same, than the results seen in
lane 3 could have been obtained by first digesting the DNA with the restriction
enzyme used in lane 1, and digesting it again with another restriction enzyme.
Answer is e)
Question 4 (2 points)
Your mother, a molecular biology expert, has brought home a photocopy of something
she is working on (illustrated below. Knowing that you are studying Basic Genetics, she
has told you that it represents a VNTR analysis of putative family members.
Please read carefully the following statements, all based on the photocopy above:
1) Individual analysed in lane 1 could be the child of individuals analyzed in lanes 3
and 4.
2) Individual analysed in lane 2 could be the child of individuals analyzed in lanes 3
and 4.
3) Individual analysed in lane 7 could be the child of individuals analyzed in lanes 4
and 5.
4) Individual analysed in lane 6 could be the child of individuals analyzed in lanes 2
and 7.
5) Individual analysed in lane 5 could be the child of individuals analyzed in lanes 1
and 3.
6) Individual analysed in lane 3 could be the child of individuals analyzed in lanes 1
and 2.
Now choose the best answer from below:
a) Only statement 5) is false, all others are true.
b) Statements 1), 2), 3) and 6) are true, while statements 4, and 5) are false.
c) All statements are true.
d) Statements 1), 2) and 6) are true, while statements 3), 4) and 5) are false.
e) Statements 3) and 5) are false, while statements 1), 2), 4) and 6) are true.
Answer is b)