Download Practice exam 1 key

Name____________________________________
Page 1 of 6
Short answer (5 questions @ 2 pts each)
1a) __AA × aa_(no partial credit)__ For the genotypes AA, Aa and aa, what cross yields
all Aa progeny?
b) ___Aa × Aa_(no partial credit)__ For the genotypes AA, Aa and aa, what cross yields
AA, Aa and aa progeny?
c) ___1:2:1 (no partial credit)_____ In Problem 2, what is the ratio of AA : Aa : aa
progeny?
d) __1x(1/2)x(1/2)x1x(1/2) = 1/8___ An individual of genotype AA Bb Cc DD Ee is
(no partial credit)
testcrossed. Assuming that the loci undergo
independent assortment, what fraction of the
progeny are expected to have the genotype Aa Bb Cc
Dd Ee?
e) ____anaphase (2 pts)__________ Stage of mitosis or meiosis in which
anaphase I (1 pt)
chromosomes oriented on opposite sides of the
metaphase plate undergo separation.
2. (10 points) There is a strain of D. melanogaster that is pure-breeding for white eyes and
another strain that is pure-breeding for red eyes. When white-eyed females are crossed to redeyed males, the progeny are red-eyed females and white-eyed males.
a) Explain this result. Give the genotypes of the parents and progeny and define all symbols
used. (10 pts)
Xw = white eye allele on X chromosome is recessive
X+ = red eye allele on X chromosome is dominant
Y = Y chromosome
P0:
F1:
XwXw female x
Xw
X+Y male
X+
Y
XwX+
red female
XwY
white male
XwX+
XwY
red female
white male
Initial description must include symbols (1 pt) and fact that white is recessive (1 pt). Correct
genotypes of parents (2 pts). Correct genotypes (6 pts) of F1 (full credit if male shown with 1
X, but no Y). (-1 to -2) if genotypes shown, but not which male and which female. If
postulated the reverse (female heterogametic, white dominant), full credit if adequately
explained and internally consistent (would also have to postulate in part b that ND occured in
male, which no one did).
Xw
Name____________________________________
Page 2 of 6
3. (20 points) Type 1 albinism in humans is a rare hereditary condition associated with a reduced
amount of tryosinase, an enzyme required for the conversion of the amino acid tyrosine to the
dark pigment melanin. In the following pedigree, males are indicated by squares, females by
circles and individuals with type 1 albinism are indicated by shading.
A
B
C
D
E
a) What is the mode of inheritance of type 1 albinism? Explain your reasoning.
(5 pts) Recessive (1 pt) because unaffected parents have affected progeny (1 pt). Autosomal (1
pt) since unaffected father has affected daughter (2 pt); affects both males and females (1 pt).
b) List the genotypes of the indicated family members as specifically as possible with the data
available. If there is more than one possibility, clearly indicate all possibilities. Except where
necessary to explain the pedigree, assume that the individuals marrying into the family are
homozygous wild-type.
(5 pts) 1 pt each
Individual A Aa or AA (or A-)
Individual B Aa or AA (or A-)
Individual C Aa
Individual D Aa
Individual E Aa or AA (or A-)
c) If individuals B and E marry, what is the probability that their first child will be affected by
type 1 albinism? Show your work.
(5 pts)
B’s father is Aa, mother is AA
probability B is Aa = 1/2 (2pts)
Name____________________________________
Page 3 of 6
E’s parents are BOTH Aa.
E is NOT aa
probability E is Aa = 2/3 (2pts)
probability that a child of 2 carriers is affected = 1/4 (1 pt)
Probability of B being carrier, E being carrier, and producing and affected child =
(1/2)x(2/3)x(1/4) = 2/24 = 1/12
-1 if equation incorrect; -1 for math error.
d) If their first child is affected, what is the probability that their second child will be unaffected?
Show your work.
(5 pts) Both B and E are carriers (2 pts); child has 3/4 chance of being unaffected (3 pts).
Child has 1/4 chance of being affected (2 pts if specify 'affected child'; 1 pt if not). If missed
the point that B and E are carriers, but correctly determined probability as if that were not
known (1 - answer in c) (1 pt).
4. (24 points) Female Drosophila melanogaster heterozygous for three recessive alleles a, b, and
c are crossed to abc males. The phenotypes of 10,000 progeny are scored as follows:
+++
4000
abc
4020
+bc
750
a++
730
+b+
230
a+c
250
++c
8
ab+
12
a) What results would you expect if a, b, and c were unlinked?
(4 pts) All 8 phenotypic classes would appear at approximately equal frequency -or1:1:1:1:1:1:1:1 -or- 1250 of each class.
b) Based on the observed results, what is the map order? Explain your reasoning.
(4 pts) A-C-B -or- B-C-A (1 pt) Least common genotypes represent double-crossover classes
(1 pt); most common represent parental classes (1 pt); gene that differs is in the middle (1 pt).
Full credit if calculated map distances to determine.
Name____________________________________
Page 4 of 6
c) Using all of the relevant data, calculate each of the three two-factor recombination
frequencies. Show your work. Express all frequencies as percentages.
(6 pts)
Recombination between
ac
cb
+++
abc
+bc
a++
+b+
a+c
++c
ab+
4000
4020
750
730
230
250
8
12
750
730
8
12
1500
230
250
8
12
500
ab
750
730
230
250
*double recombinant
2(8)*
count 2X for outside
2(12)*
genes
2000
(may also add values for ac and cb)
recomb. freq. between a + c = Rac = 1500/10,000 = 15% (2 pts); if missed DCO's (1 pt)
recomb. freq. between c + b = Rcb = 500/10,000 = 5% (2 pts); if missed DCO's (1 pt)
recomb. freq. between a + b = Rab = 2000/10,000 = 20%(2 pts); if missed DCO's (1 pt)
d) What is the frequency of double recombination? Show your work.
(5 pts)
Double recombinants are:
++c 8
ab+ 12
Frequency = (8+12)/10,000 = 20/10,000 = 0.2%
If used 2 x DCO's and calculated 0.4% (2 pts); if calculated predicted rather than observed
(0.75%, answer for next part) (1 pt).
f) What frequency of double recombination would you calculate if the data contained no
evidence of interference? Show your work
(5 pts)
Rac x Rcb = 15% x 5% = 0.75%
Use of absolute number '75' (3 pts); definition of interference (1 pt).
Name____________________________________
Page 5 of 6
Question 1 (24 pts)
6 pts A) Draw and label the leading and lagging strands on this diagram. Be sure to label the 5´
and 3´ ends of both strands.
2 pts B) Indicate where helicase would be found and its direction of movement.
4 pts C) Indicate the one or more sites where DNA polymerase would be found and its/their
direction of movement.
8 pts D)
i)
ii)
iii)
iv)
yes
no
no
no
yes
yes
no
yes
4 pts E) Remember....the primer is RNA so 3´ UUAAGCAUAU 5´
6a. If you place a culture of eukaryotic cells into medium containing 3H-thymine just prior to S,
how will this radioactivity be distributed in a pair of homologous chromosomes at metaphase?
Would the radioactivity be in (a) one chromatid of one homolog, (b) both chromatids of one
homolog, (c) one chromatid each of both homologs, (d) both chromatids of both homologs, or (e)
some other pattern? Choose the correct answer and explain your reasoning.
(d) both chromatids of both homologs — replication is semi-conservative so one strand of each
chromatid will be "new" and radioactive.
(2 pts): d (-1 if right answer for wrong reason)
(4 pts): explanation or diagram; for full credit must mention that one
strand of each chromatid is radioactive.
Partial credit (2 pts) if answered 'c' because confused chromatids and
DNA strands, if provided consistent explanation.
b. If these same cells are then allowed to go through a second S phase but with only nonradioactive nucleotides available, how would you expect the radioactivity to now be distributed
in a pair of homologous chromosomes at metaphase? (Ignore the effect recombination might
Name____________________________________
Page 6 of 6
have on this outcome.) Would the radioactivity be in (a) one chromatid of one homolog, (b) both
chromatids of one homolog, (c) one chromatid each of both homologs, (d) both chromatids of
both homologs, or (e) some other pattern? Choose the correct answer and explain your reasoning.
(c) one chromatid of both homologs — after the first division, each chromosome has one
strand that is radioactive and one that is not. When DNA replication occurs again, one
chromatid gets this radioactive strand as its "old" strand while the other chromatid gets
the non-radioactive strand as its "old" strand. The new strands are not radioactive so the
result is (c).
(2 pts): c (-1 if right answer for wrong reason)
(4 pts): explanation or diagram
Partial credit (1 pts) if answered 'e' or 'a' because confused chromatids
and DNA strands, if provided consistent explanation