BISC403
Genetic and Evolutionary Biology
Spring, 2011
April 19, 2011
Summary of requirements for Exam 3 (to be given on April 26 plus third exam from fall,
2010)
The primary responsibility is for any topic covered in lecture. This will include those
portions of the textbook which relate to the lecture.
Reading:
1.
2.
3.
4.
5.
6.
Chapters 6 and 8 as emphasized in the Molecular Biology supplement.
Mutant genes and altered proteins, section 7.3
Genetics of organelles, section 14.5, 14.6, 14.7
Bacteriophage genetics, genetic fine structure, pages 281-224
Transformation, pages 486-488 (addition to schedule of lectures)
Transduction, pages 492-494
Portions of exams 3 and 4 from fall, 2010 appear below.
QUESTIONS 16 THROUGH 30 FROM EXAM 3 OF
FALL, 2010
16)
a)
b)
c)
d)
“Inborn errors of metabolism” in humans:
describe serious diseases caused by genetic defects.
result in distinct phenotypes.
usually involve the inability to metabolize excess phenylalanine.
are now corrected by repairing the defective gene.
17)
a)
b)
c)
d)
18)
The "one gene - one enzyme" hypothesis:
was developed so that treatment could be offered for human diseases known as
“inborn errors of metabolism".
was developed by studying mutants of the fungus Neurospora unable to
synthesize arginine.
does not adequately describe the synthesis of multimeric proteins.
b) and c)
Five E. coli strains were isolated which are auxotrophs for thymine. They are
identified here by mutation number. Each was tested separately for growth on
four known precursors of thymine. A “+” indicates growth of that mutant strain in
the presence of the specific added molecule. A “-“ indicates failure of the mutant
to grow when that molecule was added.
Mutant
9
10
14
18
21
a)
b)
c)
d)
e)
A
+
+
+
-
B
+
+
-
Precursor/product
.
C
D
Thymine
+
+
+
+
+
+
+
+
+
+
Assume that molecule X is the known starting point of the pathway, preceding all
the molecules in the table above. This is a linear pathway. Which of the
following is true?
Mutant 18 is blocked at the first step of the pathway (conversion of X to D).
Molecule B will accumulate when mutant 9 tries to make thymine.
Mutant 21 is blocked at the last step of the pathway (conversion of C to thymine).
Mutant 14 is blocked at an earlier step in the pathway than mutant 10.
All of the above.
19)
a)
b)
c)
d)
e)
20)
a)
b)
c)
d)
e)
21)
a)
b)
c)
d)
The information in this question is taken directly from problem 7.32 in the book.
The pathways for the biosynthesis of the amino acids glutamine (gln) and proline
(pro) involve one or more common intermediates. Auxotrophic yeast mutants
numbered 1-7 are isolated that require either glutamine or proline or both amino
acids for their growth, as shown in the table below (+ means growth; - no
growth). These mutants are also tested for their ability to grow on the
intermediates A-E.
gln +
Mutant
A
B
C
D
E
gln
pro
pro
1
+
+
+
+
2
+
+
3
+
+
4
+
+
5
+
+
+
6
+
+
+
7
+
+
+
Each mutant is defective in a single step. Which of the following is true?
Compound D occurs earlier in the common part of the pathway than compound C.
Compound B is on the part of the pathway leading only to glutamine.
Compounds A and E are on the part of the pathway leading only to proline.
Compound D will accumulate in cultures of mutant 3.
All of the above
In class, we discussed the example of different arginine auxotrophs of
Neurospora, each of which was blocked in a single step of the arginine synthesis
pathway. We saw how the different strains were able to grow after the addition of
the appropriate intermediates and also discussed the accumulation of different
pathway intermediates in cultures of the different strains. Which of the following
is true?
The addition of an intermediate which occurs later in the pathway than the
blockage will overcome the blockage and arginine will be produced.
When a pathway is blocked, the intermediate that precedes the blockage will
accumulate in the growth medium.
The intermediate immediately following the blockage will accumulate in the
growth medium.
If arginine is added to the growth medium, then no intermediates will accumulate
since the cells are making the product they need.
all but c).
The electrophoretic mobility of a protein:
will be altered if amino acid changes lead to a change in the net charge of the
protein.
allows a prediction of the enzyme activity of the protein if it is an enzyme.
is only a useful phenotype for non-enzyme proteins, such as hemoglobin.
all of the above.
22)
c)
d)
Electrophoresis of hemoglobin molecules from a large number of different
people:
provides a new phenotype that is not directly related to oxygen-carrying ability.
explains the sickle cell phenotype because of the electrical charge change on the
hemoglobin.
can show three different bands even in people without sickle cell anemia.
a) and c).
23)
a)
b)
c)
d)
The original observations of maternal inheritance:
were made with flower color in the 4 o’clock plant.
are explained by the presence of defective mitochondria in some leaf cells.
demonstrate the variability of chloroplast genomes in plant cells.
all of the above.
24)
What is the difference between sex-linked inheritance of recessive characteristics
and maternal inheritance? Assume the absence of aneuploidy in this question.
In sex-linked inheritance, all sons have the same phenotype as their mother, but
daughters do not.
In maternal inheritance, all sons have the same phenotype as their mother, but
daughters do not.
In maternal inheritance, all offspring have the mothers’ phenotype.
For phenotypes caused by sex-linked inheritance, all offspring have the mothers’
phenotype.
a) and c).
a)
b)
a)
b)
c)
d)
e)
25)
a)
b)
c)
d)
e)
Which statements support the endosymbiont theory for the evolution of
eucaryotes?
mitochondria and chloroplasts are genetically very similar and can exchange
genes with each other in a plant cell.
The existence of a slightly altered genetic code in some mitochondria or
chloroplasts implies independent origin from the nucleus.
In humans, the division of the cell and replication of mitochondria are carefully
coordinated, indicating the mitochondria are derived from the nucleus.
Some enzymes, such as cytochrome c oxidase, have some of their subunits
determined by nuclear genes and some by mitochondrial genes.
b) and d)
26)
a)
b)
c)
d)
The amount of DNA in mitochondria in humans:
may vary in different tissues as deletions accumulate during aging.
must not change during aging or damaging phenotypes will arise.
is strongly controlled by nuclear genes.
shows strong gender relation as males inherit more of their mitochondria from
their mothers.
27)
a)
Mitochondria:
never fuse and always remain separate, dividing by binary fission.
b)
c)
d)
have multiple different circular chromosomes called nucleoids.
replicate mostly independently (semi-autonomously) of DNA synthesis in the
nucleus.
only divide after receiving instructions from genes in the nucleus.
28)
a)
b)
c)
d)
RNA editing:
is the same thing as the removal of introns from the transcript of a nuclear gene.
allows the careful cutting of a primary transcript to produce a mature mRNA.
is the formation of “guide RNA” from the information in messenger RNA.
occurs with the maxicircles and minicircles in the kinetoplast of some protozoans.
29)
Aging in humans is correlated with deletions of a few kb in the mtDNA of some
mitochondria of heart cells and brain cells. Which of the following is true?
The mtDNA with deletions are replicated more rapidly because they are shorter
and therefore become a progressively larger percentage of the total.
Mitochondria with the deletions reproduce more slowly and are continuously
removed from the cells as they are outcompeted by those with no deletions.
The mutation rate of mtDNA is much higher than that of nDNA and the deletions
protect the mitochondria by removing the damaged portions.
Replacement of mitochondria that have the deletions with mitochondria from
younger cells with intact mtDNA has been an effective treatment for Alzheimer’s
Disease.
a)
b)
c)
d)
30)
a)
b)
c)
d)
MERRF (myoclonic epilepsy and ragged red fiber disease) in humans is caused
by mitochondria with a mutated tRNA gene. Humans with this disease:
produce ATP at normal rates, but are blocked from releasing it to the cytoplasm.
form proteins with incorrect AA sequence and therefore diminished activity.
have elevated oxidative activity with excessive production of free radicals.
only show symptoms if all the mitochondria in a specific tissue are affected.
FIRST 19 QUESTIONS FROM FINAL EXAM OF FALL,
2010
1.
a)
b)
c)
d)
The plaque assay:
requires bacteria that have ceased growing so that phages can easily enter them.
begins by mixing a large number of bacteria with a small number of phages.
uses starving bacteria because they are easier for the phage to attack.
is only possible with phages that can obtain nutrients directly from the agar.
2.
a)
d)
The lytic cycle of T4 bacteriophage:
gets its name from the fact that the phage must be lysed to start the process so that its
DNA can enter the cell
has as an early step the degradation of the DNA of the bacterial host.
only begins after multiple phages have infected the host; a single phage is not enough to
initiate the process.
involves the destruction of competing phage DNA by the phage which began the cycle.
3.
a)
b)
c)
Benzer discovered that rII- T4 phages:
formed plaques that had fuzzy edges in comparison to the wild type.
formed very small plaques since their mutations slowed their growth.
could only grow in E. coli strain B.
b)
c)
d)
only formed new phage after recombination or complementation.
4.
Assume you are studying the rII region of phage T4. Four rII- strains are obtained,
identified as A, B, C and D. Each is a point mutation. When mixed infections are
performed in E. coli strain K(λ), the following results are obtained:
Strains in mixture
A and B
A and C
A and D
B and C
B and D
C and D
a)
b)
c)
d)
Result
No plaques are formed
No plaques are formed
Plaques are formed
No plaques are formed
Plaques are formed
Plaques are formed
The conclusion most consistent with these results is:
A and C have mutations in the same gene, B and D are in a different gene.
A and B have mutations in the same gene, C and D are in a different gene.
A, C and D have mutations in the same gene, B is in a different gene.
A, B, can C have mutations in the same gene, D is in a different gene.
5.
Consider an experiment similar to that described in question 4. Once again you have four
rII- mutants, but this time they are all deletion mutations. If you did the same type of
experiment, namely mixed infections of E. coli strain K(λ), with all possible pairs of
mutants, which of the following would you expect?
a)
b)
c)
d)
plaques would never be formed, since deletions cannot complement each other.
if the two deletions were in the same gene but did not overlap, plaques would be formed.
if the two deletions were in different genes, plaques would not be formed.
none of the above.
6.
Shown below are the deletion maps of a series of rII- mutations. The deleted region is
indicated as (….) and the intact region as _____. Note that strain 5 carries two different
deletions.
1 _____(……..)______________________
2 ________(……..)___________________
3 ____________(……..)_______________
4 ________________(……..)___________
5 ___(.......)____________(……..)_______
A series of point mutations A-E was used in mixed infection experiments, that is the
coinfection of the proper E. coli strain by one point mutation and one deletion. The
results of those experiments are shown below. The ability to produce wild-type progeny
phage is shown by (+) and (0) indicates that no wild-type progeny phage were produced.
Note that strain 5 carries two different deletions.
A
B
C
D
E
1
+
0
+
0
+
2
+
+
+
0
+
3
+
+
0
+
+
4
0
+
+
+
0
5
0
0
+
+
+
Indicate the order that is most consistent with these data.
a. DCEAB
b. ABCDE
c. BDCEA
d. AECBD
e. CEABD
7.
a)
b)
c)
d)
Benzer's recombination experiments that were performed after deletion mapping:
required a two stage experiment, with growth in E. coli strain K(λ) first.
resulted in the production of rII- mutants from rII+ wild type phage.
determined the exact location of even the most closely linked mutations.
All of the above.
8.
a)
b)
c)
Recombination in the rII region of phage T4:
may be detected between two deletions as well as between two point mutations.
was only observed when one of the mutations was a deletion.
produced only wild type phage chromosomes since the mutations had been removed.
d)
could not have been detected if the two mutants being tested had defects in different
genes.
9.
a)
b)
c)
d)
Benzer found "hot spots" for point mutations in the rII region of phage T4 which may be
explained as:
parts of the chromosome in which recombination happens more frequently.
parts of the chromosome in which the effects of mutation are seen more readily.
the formation of deletions which cover the positions of large number of rII mutants.
either a or b.
10.
a)
b)
c)
d)
Transduction:
can transfer DNA from any bacterial species to any other bacterial species.
requires large phage that have room to carry bacterial DNA in addition to their own.
benefits the phage by allowing its DNA to be distributed more efficiently.
can only be detected after the recipient cell undergoes recombination.
11.
a)
b)
c)
d)
Generalized transduction:
transfers complete bacterial chromosomes.
usually transfers a different portion of bacterial DNA each time.
always involves the production of defective phages.
b and c
12.
a)
b)
c)
d)
A bacterial cell which has been transduced:
can be detected more easily if it received a recessive allele in the transfer.
remains intact after infection (does not lyse) because the transducing phage is always
defective.
can itself become a donor in a subsequent transduction.
b and c.
13.
a)
b)
c)
d)
Specialized transduction:
can only be done by virulent phages.
may transfer bacterial genes from any portion of the donor chromosome.
occurs more rapidly when there is cell to cell contact.
causes the donor bacterial cell to lyse.
14.
a)
b)
c)
d)
The integration of phage λ into the bacterial chromosome:
occurs at the same insertion sites used by the F factor in forming Hfr.
blocks further transcription of almost all bacterial genes.
leads to the formation of a repressor protein by the CI gene.
is the first step in the lytic cycle of temperate phages.
15.
a)
b)
c)
d)
The induction of a prophage:
causes the lytic cycle to begin.
results in the destruction of defective phages.
stimulates the metabolism of the infected cell.
begins with the synthesis of a repressor protein.
16.
A temperate phage that has a mutation in its CI gene so that an inactive repressor is
formed:
a)
b)
c)
d)
could still enter into a lysogenic cycle.
would form a prophage that could not be induced.
might possibly still cause transduction.
all of the above.
17.
a)
b)
c)
d)
e)
Transformation:
occurs in nature as well as in the laboratory.
can occur with genes from non-bacterial species such as humans.
can involve plasmids as well as linear portions of the donor chromosome.
of genes from the bacterial chromosome is usually detected by the transfer of dominant
alleles to recipients that had recessive alleles.
all of the above.
18.
a)
b)
c)
d)
Competence of bacteria:
describes the ability of bacteria to donate DNA in transformation.
is higher in F- cells since they are better at receiving DNA.
is the term for the transfer of DNA, from one species to another.
can be increased in the laboratory by a pulse of electricity.
19.
a)
Cotransformation:
usually involves recipient bacterial cells that are dominant for all the alleles being
transferred.
is the term which describes the transformation of two cells at once.
is rarely used to map bacterial chromosomes because the experiments are more difficult
to do than conjugation.
does not occur for genes which are very far apart.
None of the above.
b)
c)
d)
e)
Answers:
Exam 3 answers
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
b
d
e
e
e
a
d
c
e
e
a
c
d
a
b
Exam 4 answers
1. b
2. b
3. c
4. d
5. b
6. c
7. c
8. a
9. b
10. d
11. d
12. d
13. d
14. c
15. a
16. c
17. e
18. d
19. d
Short answer questions from exam 3
32.
a)
b)
In corn snakes, the wild type color is brown. One autosomal recessive mutation
causes the snake to be orange and another causes the snake to be black. An orange
snake was crossed to a black one, and the F1 offspring were all brown. Assume
that all relevant genes are unlinked. Answer the following two questions.
What phenotypes and in what ratios would you expect in the F2 created by mating
two of the F1 snakes assuming that there is a single pigment pathway, with the
orange and black colors coming from the accumulation of different intermediates
in the pathway that makes brown pigment?
What phenotypes and in what ratios would you expect in the F2 created by mating
two of the F1 snakes assuming that there are two different pigment pathways,
with orange pigment the product of one and black pigment the product of the
other? In this case brown is the result of mixing the orange and black pigments in
the skin of the snake.
Answer:. This question was copied directly from problem 7.28 in the textbook. The
essential features of the two answers were:
a)
This is a description of complementary gene action, directly parallel to the
production of purple flowers from white-flowered parents in sweat peas. The
expected ratios are: 9 brown: 4 orange: 3 black or 9 brown: 4 black: 3 orange,
depending on which color is produced first in the pathway.
b)
Since these are two unlinked genes operating on separate pathways, the
results are: 9 brown: 3 black: 3 orange: 1 white. Some students wrote that the
last class was “undetermined” and this answer received credit as long as it was
clear it was not brown, orange or black.
33.
Consider the following three pedigrees for a trait known to be transmitted by
maternal inheritance.
Pedigree 1
Answer:
Pedigree 2
Pedigree 3
Pedigree 1 is b
Pedigree 2 is c
Pedigree 3 is a
Short answer question from exam 4
31.
Benzer’s complementation experiments were done with E. coli strain K(λ),
whereas the recombination experiments began with E. coli strain B. Explain.