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INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
Thursday, October 28, 2010
STUDENT NAME: _______________________________________________
(Last)
(First)
Circle TA name
Chaochao
Diana
Edwin
ID # _________________
Jenny
Jessica
Paige
INSTRUCTIONS
1. This exam is 14 pages long including this front page. Before starting make sure you have all the pages. Put your name
and ID on each page!!!!!!!!!
2.
No credit will be given for an illegible answer. Regrades will only be considered for answers written in pen (not pencil).
3.
No graphing calculators are allowed.
4.
Show all your work!! You will not be given credit for answers alone. Circle your answers!
5.
Go forth and do genetics!
QUESTION
VALUE
1
6
2
8
3
10
4
15
5
8
6
12
7
18
8
8
9
16
10
14
11
12
12
8
13
15
SCORE
TOTALS 150
1
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(1) (6 points) Chaochao A pea plant from a pure-breeding strain that is tall with yellow pods is crossed
to a plant from a pure-breeding strain that is dwarf with green pods. The F1 plants are all tall and have
green pods. (Assume complete dominance and no linkage).
a. (2 points) How many genetically distinct gametes can be produced by the F1?
4 – TG, Tg, tG, tg
b. (2 points) What phenotypes and in what ratios do you expect to see in the F2?
Tall green 9/16
Tall yellow 3/16
Dwarf green 3/16
Dwarf yellow 1/16
c. (2 points) If you crossed an F1 plant back to the pure breeding parent that is dwarf with green
pods, what phenotypes and in what ratios do you expect to see in the progeny?
Tall green 1/2
Dwarf green 1/2
2
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(2) (8 points) Chaochao A particular trait is present in these two families (the same trait in both).
a. (2 points) What is the most likely mode of inheritance for this trait? Be very specific and explain
your answer.
Autosomal Dominant. Two affected parents having an unaffected child. Can be passed from father to
son, so most likely not X linked.
b. (1 point) What is the genotype for II-2? (use A for dominant allele and a for recessive allele)
Aa
c. (1 point) What is the genotype for II-10? (use A for dominant allele and a for recessive allele)
aa
d. (4 points) If III-4 married III-5, what is the probability that their first child will have this disease?
1/3*1 +2/3*0.5=2/3
3
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(3) (10 points) Paige Consider the following pedigree for a rare genetic disease:
a) (2 points) What is the most likely mode of inheritance? Give 2 reasons to support your hypothesis.
Autosomal recessive; Males and females are equally likely to be affected; Approximately ¼ of progeny
from carrier parents will be affected; an affected child has 2 unaffected parents (carriers); trait skips
generations
b) (2 points) Which nonexpressing members of this family must carry the mutant allele?
I-1, I-2
II-5
III-1, III-2, III-5
IV-1, IV-2
c) (2 points) If individual II-3 married an outsider, what is the probability that their first child would be a
carrier?
P(II-3 Aa)=2/3. P(child inherits a) = ½, therefore (2/3)*(1/2) = 1/3 chance child will be a carrier.
d) (4 points) If IV-1 and IV-2 have four more children, what is the chance that exactly 2 of them will be
affected?
6 X ¼ X ¼ X ¾ X ¾ = 27/128.
4
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
Thursday, October 28, 2010
STUDENT NAME: _______________________________________________
(Last)
(First)
ID # _________________
(4) (15 points) Jessica A human pedigree is shown below. The ABO blood types of some of the
members of the family are shown.
a.
(8 points) For the individuals listed below, circle their correct genotype. If there is more than one
possible genotype, then circle all of the possibilities.
Individual
b.
Possible Genotypes
I-1
IA IA
IA IB
IB IB
IA i
IB i
ii
II-2
IA IA
IA IB
IB IB
IA i
IB i
ii
III-2
IA IA
IA IB
IB IB
IA i
IB i
ii
IV-1
IA IA
IA IB
IB IB
IA i
IB i
ii
(7 points) What is the probability that IV-1 will have blood type AB? (circle the correct answer)
1 15/16 7/8 3/4 5/8 9/16 1/2 7/16 3/8 5/16 1/4 3/16 1/8 3/32 1/16 1/32 1/64 0 (none of above)
(You must show your work for credit or partial credit)
For II-2 Bi is ½ chance and AB is ½. Therefore, for III-2, Ai = ¼, Bi = ½, ii = ¼.
B from III2 x A from III-3 is (1/2 x 1/2) x (1/2 x 1/2) = 1/16
+ A from III2 x B from III-3 is (1/4 x 1/2) x (1/2 x 1/2) = 1/32
5
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(5) (8 points) Edwin Color in frogs is controlled by the C gene, and you know that frogs with genotype
CB CB are brown, frogs with genotype CG CG are green, and frogs with genotype CG CB are green. Your
friend comes back from Costa Rica with two bright red frogs: CR –.
When he breeds the original red frogs with the CG CB green frog, he gets 29 offspring:
14 red frogs
8 brown frogs
7 green frogs
a) (5 points) What is the genotype of his original red frogs?
CR CB
He decides he wants a true-breeding strain, so he crosses his original two red frogs together. He gets
some red and some brown frogs. He then crosses several pairs of F1 red frogs together but fails to get
a true breeding line. In total, he generates several dozen offspring with the following phenotypes:
36 red frogs, 17 brown frogs.
b) (3 points) What is the most likely explanation for this and how do you know?
CR CR is LETHAL. Frogs dying, seeing a 2:1 ratio.
6
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(6) (12 points) Jessica Zebrafish are small fish with 5 dark stripes that run the length of their bodies.
Zebrafish are very important for biological research because they are transparent for the first week of
life. To further help scientists, you are attempting to make true breeding lines of zebrafish that stay
transparent as adults. At one pet shop you find a transparent female zebrafish. You cross the
transparent zebrafish female with a true breeding wild type male fish and the F1 generation fish are
completely normal zebrafish with stripes. When you cross two F1 fish, you get 746 wild type fish and
54 transparent adult fish.
(a) (4 points) What is the mode of inheritance for this trait in zebrafish?
Duplicate genes
(b) (2 points) What is the genotype of the fish you found at the pet shop?
aabb
(c) (2 points) What is the genotype of the F1 fish?
AaBb
(d) (4 points) It turns out that transparent male fish are sterile, so to generate large numbers of
transparent fish, you cross the F1 male fish back to the original female fish you found at the pet shop.
What proportion of the fish from this cross would be transparent. Show your work.
Male AaBb x female aabb → AaBb, Aabb. aaBb, aabb = 1/4
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INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(7) (18 points) Paige Chromosome analysis of W. newlus, a newly discovered frog, indicates that the
species is diploid with two pairs of chromosomes (autosomes), one long and one short. Simple genetic
analysis indicates that the gene (A) that specifies amber spots is located on the long chromosome, and
a gene (B) that specifies body width resides on the short chromosome.
(a) (6 points) Diagram the appearance of the chromosomes at metaphase of meiosis I for a W. newlus
cell that is heterozygous for both genes. You must specify all chomatids of all chromosomes and all
copies of alleles.
A
X A aX a
B B b b
x
x
(b) (12 points) Do the following occur in meiosis I, meiosis II or mitosis? (Each blank may contain more
than one answer)
i) At least one crossover per homologous pair can occur
______________________________
ii) Chromosomes line up singly along the metaphase plate
______________________________
iii) Produces somatic cells
______________________________
iv) Homologous chromosomes separate from each other
______________________________
v) Ploidy level (n) is decreased by half
______________________________
vi) DNA synthesis takes place prior to division
______________________________
i) Meiosis I
ii) Mitosis, Meiosis II iii) Mitosis iv) Meiosis I v) Meiosis I vi) Mitosis, Meiosis I
8
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(8) (8 points) Jenny You’ve discovered a new dominant X-linked mutation in humans that confers the
ability to teleport. A teleporting woman whose mother cannot teleport marries a man who can also
teleport. Their son has Klinefelter syndrome (XXY) and cannot teleport.
a) (4 points) In which parent did non-disjunction occur? or can it not be determined? Draw the
pedigree and explain your reasoning.
Non-disjunction occurred in the mother
b) (4 points) Did non-disjunction occur in meiosis I or meiosis II, or can it not be determined? Diagram
the non-disjunction for full credit.
Meiosis II.
9
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(9) (16 points) Edwin A breed of hamsters has color controlled by two genes, denoted S for ‘sand
colored’ and G for ‘gray colored,’ with double-recessive hamsters being albino white. The pet store
crosses true breeding sand-colored hamsters with true breeding gray-colored hamsters. All of the F1
progeny turn out to be sand colored. When sets of F1 progeny were mated, the following F2
phenotypes were observed:
116
33
11
sand colored hamsters
gray colored hamsters
pure white hamsters
a) (3 points) What is the mode of gene interaction?
DOMINANT EPISTASIS of sand (S) to gray (G or g).
b) (3 points) Assuming the S and G genes assort independently (no linkage), what are the expected
proportions of hamster colors in the F2
12:3:1
Sand: gray: white
c) (10 points) Do the results of the cross differ significantly from expected results: Support your answer
by showing your complete work, show the chi square value, the p value, and state the null hypothesis.
Observed
Sand 116
Gray 33
White 11
Sum:
160
Expected
12/16 *160
3/16 * 160
1/16 * 160
E
120
30
10
(0-E)
4
3
1
Sum:
|O-E| ^2 / e
.1333
.3
.1
0.5333 = chi square value
Degrees of freedom: 2, 0.5<p<0.9. Between 50% to 90% chance these results vary from the null
hypothesis of non-linkage by chance alone. Can accept null hypothesis of independent assortment.
10
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
Thursday, October 28, 2010
STUDENT NAME: _______________________________________________
(Last)
(First)
ID # _________________
(10) (14 points) Paige In Drosophila, three autosomal recessive genes have the following map:
a
20 mu
b
15 mu
c
A triply mutant female fly is crossed with a wild type male. Resulting F1 females heterozygous for each
of the three recessive mutations are then test crossed to males showing all three mutant phenotypes.
(a) (2 points) How many gametes do you expect the F1 female to produce? How many gametes do
you expect the test cross male to produce?
Female: 23 = 8
Male: 1 gamete abc
(b) (10 points) If you analyze 1000 flies from the last cross, provide the expected number of flies for
each of the following phenotypic classes. Label each class as parental (P), single recombinant (SCO),
or double recombinant (DCO):
a+b+c+
a+b+c
a+bc
a+bc+
ab+c+
ab+c
abc+
abc
P
SCO2
SCO1
DCO
SCO1
DCO
SCO2
P
340
60
85
15
85
15
60
340
(c) (2 points) If the order of genes was b-a-c instead, what would the female double recombinant
genotypes be? Draw the chromosomes and the recombinations required to produce these recombinant
genotypes:
b+ac+
ba+c
b+
a+
X
b
c+
X
a
c
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INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(11) (12 points) Diana You perform interrupted mating experiments with the following cross: Hfr strs
met+ arg+ trp+ lac+ x F- strr met- arg- trp- lac-. Gene order is not necessarily implied by the order they
are listed in. The met+, arg+, and trp+ genes are required for the biosynthesis of the amino acids
methionine, arginine, and tryptophan, respectively. The lac+ gene is required to break down the carbon
source lactose so that the cell can use it for growth. Lac- cells cannot grow when lactose is the only
carbon source, but can grow on minimal media plus glucose. You allow the bacteria to mate for certain
time intervals, and then plate the mixture on four types of media that selects either for Met+, Arg+,
Trp+, or Lac+ exconjugants. The results are summarized in the table below.
(a) (4 points) What would you add to minimal media to select for the above exconjugants:
Met+ cells?
MM + streptomycin, arginine, tryptophan, and glucose.
Arg+ cells?
MM + streptomycin, methionine, tryptophan and glucose
Trp+ cells?
MM + streptomycin, methionine, arginine and glucose
Lac+ cells?
MM + streptomycin, methionine, arginine, tryptophan and lactose (no glucose).
(b) (5 points) Based on the table, draw the best possible map for the order and approximate
distances (in minutes) of the genes in relation to each other and the origin of transfer.
<I-----lac-----arg/trp-----met-----5’
10’
15’
(c) (3 points) Name three methods you could use to resolve any ambiguities in your map.
Three-point cross, reciprocal cross, co-transduction
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INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(12) (8 points) Jenny In a conjugation experiment, you find that there are two genes that lie so close
together that you are unable to determine a gene order. There are two possible gene orders to
consider:
mal------------gal-tyr
or mal----------tyr-gal
You set up reciprocal crosses so that the Hfr donor is mal+ gal+ tyr- and the F- recipient is mal- galtyr+ in one experiment (cross A) or the Hfr donor is mal+ gal- tyr+ and the F- recipient is mal- gal+
tyr- in another experiment (cross B). You screen for mal+ gal+ tyr+ exconjugants and compare their
frequencies in both experiments, and find that cross A produces about 1/25th as many as cross B.
What is the correct gene order? You must show your work for credit.
mal tyr gal
13
INSTRUCTORS: Steve Jacobsen
Alvaro Sagasti
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 28, 2010
ID # _________________
(13) (15 points) Diana You have a donor strain of bacteria that is thr+ ala+ pro+ and a recipient strain
that is thr- ala- pro-. Thr, ala, and pro are genes encoding the amino acids threonine, alanine, and
proline. You perform a generalized transduction experiment and initially select for thr+ cells on a
master plate. From there, you replica plate onto the following plates to select for the other markers:
Master plate (MM +ala +pro)
Replica plate 1 (MM +ala)
Replica plate 2 (MM +pro)
Replica plate 2 (MM)
430
80
260
10
(a) (4 points) What are the genotypes of the colonies on each plate?
Master: thr+ ala+/- pro+/Plate 1: thr+ ala+/- pro+
Plate 2: thr+ ala+ pro+/Plate 3: thr+ ala+ pro+
(b) (4 points) Determine the number of progeny for the following genotypes:
thr+ ala+ pro+ 10
thr+ ala- pro+ 70
thr+ ala+ pro- 250
thr+ ala- pro- 100
(c) (2 points) What is the cotransduction frequency of thr and ala?
(10+250)/430 = ~60.5%
(d) (2 points) What is the cotransduction frequency of thr and pro?
(10+70)/430 = ~18.6%
(e) (3 points) which is further from thr: ala or pro? Explain why for full credit.
pro. Thr and pro are cotransduced much less than thr and ala are, indicating that thr and ala
are closer and more likely to packaged in a phage together.
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