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Biology 105 - Summer 2013
MIDTERM EXAM
Name: ___________________________________
SCORE: ____________________________________
5 X 2 Point Questions
10 points
12 X 5 Point Questions
60 points
4 X 10 Point Questions
40 points
Drop the lowest of the 10 point questions
-10 points
TOTAL POSSIBLE SCORE
100 points
2 POINT QUESTIONS
1. [2 POINTS] How many different types (genoytypes) of gamete could be formed by an individual that is Aa
BB cc Dd?
_______(2) (1) (1) (2) = 4__________
2. [2 POINTS] The map distance between gene A and gene B is 10 map units. The map distance between
gene B and gene C is 20 map units. However, in a testcross of an individual heterozygous for genes A and C
only 26% of the progeny are recombinant. Briefly explain how can you reconcile the result of the testcross
with the genetic map.
A and C must be 30 map units (cM) apart. In a cross, they exhibit only 26% recombination due to the
effect of multiple crossovers.
3. [2 POINTS] If you have a DNA molecule with 1,000 base pairs (1 kb) whose GC% value is 60%, how many
hydrogen bonds will there be between the two strands?
(600 X 3) + (400 X 2) = 2,600
4. [2 POINTS] Which of the following treatments will be likely to stabilize the DNA double helix and reduce the
conversion to single strands? Check any/all that apply.
___ raising the temperature
_X_ raising the Mg
++
concentration
_X_ increasing the %GC in the DNA sequence
_X_ removing or inactivating helicases
_X_ allowing basic proteins (histones) to bind to the DNA
5. [2 POINTS] In the Charlie Chaplain paternity case the mother was blood type A and the child was blood
type B.
a. Which blood type phenotypes are possible for the father?
_________B or
AB_______
b. What is the mother’s genotype?
_____________ I i ____________
c. What is the child’s genotype?
_____________ I i ____________
A
B
5 POINT QUESTIONS
6. [5 POINTS] The red vizcacha rat (Tympanoctomys barrerae) is a species of rodent in the family
Octodontidae. The rodent is not a rat, but related to guinea pigs and chinchillas. This species has the highest
number of chromosomes known among mammals; somatic cells have 2n = 102.
How many chromatids and chromosomes are present
#
CHROMATIDS
#
CHROMOSOMES
a. at anaphase of mitosis in an adult somatic cell
204
204
b. during G1 prior to meiosis I
102
102
c. in a sperm cell
51
51
d. during G2 of interphase in an adult somatic cell
204
102
e. in secondary oocyte
102
51
7. [5 POINTS] In Drosophila, the following pairs of alleles are known:
Wing size:
t = recessive allele for tiny wings
T dominant allele for normal wings
Eye Shape:
n = recessive allele for narrow eyes
N = dominant allele for oval (normal) eyes
For the two genetic crosses below (A. and B.) determine the genotypes of each parent by analyzing the
results of the cross. You should assume that the genes are NOT linked.
A.
PARENT 1 Phenotype
normal oval
PARENT 2 Phenotype
X
normal oval
OFFSPRING Phenotypes
900 normal oval
300 normal narrow
300 tiny oval
100 tiny narrow
PARENT A1 GENOTYPE: ________Tt Nn________
PARENT A2 GENOTYPE: ________ Tt Nn ________
B.
PARENT 1 Phenotype
tiny oval
PARENT 2 Phenotype
X
normal narrow
PARENT B1 GENOTYPE: ________tt NN ________
PARENT B2 GENOTYPE: ________ Tt nn ________
OFFSPRING Phenotypes
500 normal oval
500 tiny oval
8. [5 POINTS] In wild populations of foxes there are a few individuals with silver fur rather than the typical
reddish-brown fur. A fox rancher who sells pelts to the fashion industry realizes that silver pelts would fetch a
higher price than the ordinary brown and so initiates a breeding program to develop a pure bred line of silver
foxes. Pairs of silver foxes are trapped in the wild and mated to each other in captivity where they always
produce litters that are 1/3 brown and 2/3 silver. Thereafter, when the silver foxes born in captivity are mated
to each other they always produce litters that are 1/3 brown and 2/3 silver just as their parents did. Therefore
the rancher fails to produce a pure breeding line of silver foxes. This is surprising because the brown foxes
always breed true for brown. Another interesting feature of the mating of silver foxes was that they produced
smaller litters than matings between brown foxes.
Propose (briefly) a genetic explanation for this episode.
The allele that causes silver fur in the heterozygous condition is a recessive lethal. This leads to the
modified monohybrid ratio of 1/3 : 2/3 as follows:
P:
Ss [SILVER]
X
F1:
1/4 SS [BROWN]
1/2 Ss [SILVER]
1/4 ss [LETHAL]
Ss [SILVER]
Use your explanation to predict the result of a cross between a brown and silver fox.
Cross: SS [BROWN] X Ss [SILVER]
Predict:
1/2 SS [BROWN]
1/2 Ss [SILVER]
normal litter size
9. [5 POINTS] This problem involves 2 genes, genes A and B which are 8 map units (cM) apart on the same
chromosome.
You begin with 2 homozygous individuals with genotypes AA bb and aa BB. (a and b are recessive alleles.)
These homozygous individuals are crossed (to each other) to obtain an F1 progeny. The F1s are then
testcrossed.
What % of the progeny of this testcross progeny will be homozgous recessive for both genes?
_____4_____%
10. [5 POINTS]
In box A. insert the correct name for the BASE (not the name of the nucleotide). A. = Cytosine
In box B. correctly indicate the direction/"polarity" of the DNA strand. B. = 5' B.
A.
11. [5 POINTS]
3
A. If you expose a culture of growing cells to radioactively labeled thymidine ( H-thymidine) during S phase,
how would the radioactive label be distributed over a pair of homologous chromosomes at the next mitotic
metaphase?
The radioactive label would appear in ______d______ (Insert the best answer from the choices below.)
a. one chromatid of one homolog
b. both chromatids of one homolog
c. one chromatid of each homolog
d. both chromatids of both homologs
e. random locations
B. If you remove the labeled thymidine after S phase, and then let the cells go through another S phase in the
labeled thymidine how would it be distributed now at the following mitotic metaphase?
The radioactive label would appear in ______c______ (Insert the best answer from the choices above.)
12. [5 POINTS] Our model for DNA replication in E. coli invoked 2 different DNA Polymerase enzymes, Pol I
and Pol III. While both enzymes are capable of elongating the 3' end of a pre-existing strand, Pol I has an
important function that Pol I does not have. Describe the special role of DNA Pol I.
Pol I is a 5' to 3' exonuclease that removes the RNA primers used in dioscontinuous synthesis on the
lagging strand (i.e. from the Okazaki pieces).
13. [5 POINTS] Matching. For each listed function choose the correct E. coli protein. Write the letter that
identifies the protein in the correct box of the column provided. Each protein should be used exactly once.
Identify the
protein by
FUNCTION
PROTEINS
writing the
correct letter
F
A. Topoisomerase
5' → 3' DNA template-dependent DNA
polymerase responsible for the majority of
leading and lagging strand DNA synthesis
Seals (repairs) single-strand nicks remaining
D
B. Primase
between adjacent Okazaki fragments in
discontinuous replication on lagging strand
Stabilizes DNA single-strands/prevents them
G
C. Helicase
from re-forming double-stranded DNA
Removes ("relaxes") the positive supercoiling
A
D. DNA Ligase
introduced by helix unwinding
H
E. DNA Polymerase I
An enzyme with 3' → 5' DNA polymerase
activity (i.e. can add nucleotides to the 5' end
of a strand).
Unwinds the DNA ahead of the replication fork
C
F. DNA Polymerase III
at rates approaching 1,000 bp/second.
Requires ATP hydrolysis.
An RNA polymerase that makes the primers
B
G. Single-Strand DNA
for synthesis of Okazaki fragments
Binding Protein
E
H. NONE, does not
5' → 3' DNA template-dependent DNA
exist.
polymerase responsible for synthesizing DNA
on the lagging strand only
14. [5 POINTS]
The diagram below shows two genes, Gene A and Gene B. The complementary DNA stands are designated
Watson and Crick.
Gene A is transcribed from right to left using Watson as the template strand.
Gene B is transcribed from the promoter PB whose location is shown.
PB
Watson
Gene A
Gene B
Crick
a. Which strand (W or C) is the template for transcription of gene B? (WATSON or CRICK)
b. Label the right-hand end of the CRICK DNA strand as either 5' or 3'.
5'
15. [5 POINTS]
Describe (in as much detail as you can muster) the molecular structure of the 5' end of a mature (processed)
eukaryotic mRNA.
The 5' end has methylguanine attched via its 5' carbon (i.e. "backwards") via 3 phosphates, to the 5'
carbon of the first nucleotide in the transcript. See Fig. 8.13.
Partial credit for "methylguanine".
10 POINT QUESTIONS
16. [10 POINTS] Assume you have 2 homozygous, true-breeding, varieties of a hypothetical organism. One
variety is always RED when inbred and the other variety is always WHITE.
When you cross these varieties to each other, the F1 generation consists entirely of RED individuals. When
the RED F1's are inbred (crossed among themselves) you obtain a population of F2's with a phenotypic ratio
of 15 RED to 1 WHITE.
Work out the genetics of the inheritance of the color difference in these two varieties and predict the result of
a cross between a RED F1 individual and their WHITE homozygous parent.
The best prediction is that the phenotypic ratios in the progeny of this cross will be:
This is a dihybrid cross in which the two genes are redundant in the sense that a single dominant
allele of either gene will express the RED phenotype. WHITE is expressed only by individuals that are
homozygous recessive at both genes.
A_ B_
A_ bb
aa B_
aa bb
9/16
3/16
3/16
1/16
RED
RED
RED
WHITE
So testcrossing the heterozygous F1s gives 3/4 RED and 1/4 WHITE.
17. [10 POINTS] In this problem assume the following:
There is a gene affecting height. Expression of a dominant allele (H) gives tall and expression of the recessive
allele (h) gives short.
There is a second gene affecting girth. Expression of a dominant allele gives thick (G) while expression of the
recessive allele (g) gives thin.
You locate a tall, thick individual in nature and in order to determine the genotype of this individual you
perform a testcross. The results of the testcross are:
501
499
tall,
short,
thick
thick
Next, you take a tall, thick individual from the progeny of the first testcross and use them to perform a second
testcross. The results of the second testcross are:
450
450
50
50
tall,
short,
tall,
short,
thick
thin
thin
thick
Explain these genetic results as fully as possible. In other words, what information can you deduce from the
results of the 2 testcrosses?
1. The tall, thick individual from nature was Hh GG.
2. The genes are linked and 10 map units (cM) apart.
18. [10 POINTS] MATCHING
COLUMN A
Physical pairing of homologous chromosomes during Meiosis I.
Sequence recognized by an RNA polymerase as a transcription
initiation site
Alternate forms of a single gene
Alleles of one gene mask or modify the expression of alleles of
a different gene
Functionally inactivated X chromosome in cells of mammalian
females
Subunits (components) of spliceosomes
The allele expressed in the phenotype of a heterozygous
individual
Sequences removed from an RNA transcript before translation
Process that modifies the 3' end of eukaryotic pre-mRNA
transcripts
A nitrogenous base containing a double ring found in nucleic
acids
A nucleotide in DNA containing a single ring in the base.
A genetic cross in which one of the two parents is known to be
homozygous recessive
One gene affecting more than one phenotypic character
i
m
COLUMN B
a. epistasis
b. Barr Body
h
a
c. dominant
d. introns
b
e. testcross
l
c
f. poly A tailing
g. purine
d
f
h. alleles
i. synapsis
g
j. thymidine
j
e
k. pleiotrpy
l. snRNPs
k
m .promoter
19. [10 POINTS] Red-green colorblindness in humans is due to the expression of a recessive allele of an Xlinked gene. A woman with normal vision whose mother was color-blind has a child fathered by a color-blind
man. What is the probability that their first child will be color blind?
C
c
The mother must be heterozygous (X X ). So 1/2 the offspring would be colorblind.
20. [10 POINTS] The pedigree diagram below shows the inheritance of a rare liver disease.
A. What is the most likely genetic basis for the inheritance of this disease (check one option below)?
4 pts.
autosomal recessive
autosomal dominant
X-linked recessive
X-linked dominant
__X___
_____
_____
_____
B. If individuals 1 and 2 have a child, what is the probability that their first child will have the disease?
6 pts.
Since this is a RARE condition we can assume that the exogamous individuals (those that marry into
the family) do not carry the allele responsible (they are homozygous for the normal dominant allele).
In this case the probability that individual 1 inherits the recessive allele is 1/2. On the other hand,
individual 2 must be heterozygous because one of their parents was homozygous.
So the probability that a child of 1 X 2 is affected would be (1/2) (1) (1/4) = 1/8