Download Midterm Key

INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
Thursday, October 31, 2013
STUDENT NAME: _______________________________________________
(Last)
(First)
Circle TA name
Ao
Bella
Linda
Manon
Josh
ID # _________________
Kevin
Michael
INSTRUCTIONS
1. This exam is 15 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! Use the back for
scratch paper only, we will not be grading the back.
QUESTION
VALUE
SCORE
TA
1
12
Josh
2
15
Manon
3
12
Kevin
4
15
Michael
5
12
Bella
6
8
Manon
7
10
Kevin
8
12
Ao
9
15
Josh
10
14
Bella
11
11
Linda
12
14
Josh
TOTALS 150
1
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
(1) Josh (12 pts) Genes G and H show independent assortment and each gene has two alleles with one allele showing
complete dominance over the other.
a) (6 pts) You cross a plant with genotype GgHh to a plant with genotype ggHh. What fraction of the progeny do you expect
to show at least one dominant trait?
7/8
P(gghh) = 1/2 x 1/4 = 1/8
P(at least one dominant trait) = 1 – 1/8 = 7/8
b) (6 pts) You self-cross a plant with genotype GgHh. What fraction of the progeny do you expect to be true-breeding
(homozygous)?
1/4
P(GGHH or GGhh or ggHH or gghh) = 4/16 = ¼
2
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
Thursday, October 31, 2013
STUDENT NAME: _______________________________________________
(Last)
(First)
ID # _________________
(2) Manon (15 pts) Henry and Lea are expecting a child. Both are healthy adults, but Lea’s mother and Lea’s sister Kate
suffer from myotonic dystrophy. Lea’s father is unaffected. Kate’s husband John is also affected, and their son is healthy.
Henry’s father suffers from myotonic dystrophy, but Henry’s sister and mother are healthy.
a)
(6 pts) What is the likely mode of inheritance for myotonic dystrophy. Be specific and include a drawing of the
pedegree?
Autosomal Dominant
b) (3 pts) What is the probability that Henry and Lea’s child will have myotonic dystrophy?
0. Both parents are homozygous wild type
c)
(3 pts) If Kate and John have another child, what is the probability that the child will be an affected girl?
0.75 x 0.5 = 3/8
d) (3 pts) Kate and John want 3 additional children. What is the probability that only 1 out of the three children will be
affected?
(0.75)(0.25)(0.25)(3) = 9/64
3
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
(3) Kevin (12 pts) ABO blood groups. Blood types A, B, and O are determined by the presence of antigens, IA for A, IB for
B, and i, a lack of antigen, for O
Harold has blood type B, one of his parents has blood type A and the other has blood type B. His brother has blood type AB.
Harriet has blood type A, and her parents both have blood type A. Her sister has blood type O.
a) (6 pts) What is the probability that Harold and Harriet’s first child will have blood type AB?
Harold has genotype IB i (must have received i allele from parent with type A).
Harriet either has genotype IA i (2/3 chance) or IA IA (1/3 chance). (Her parents are both genotype IA i.)
Chance for AB = 1/2 (chance of IB allele from Harold) x [(2/3 x 1/2) + [1/3 x 1)] (chance of IA allele from Harriet)
= 1/2 x 2/3 = 1/3
b) (6 pts) What is the probability that Harold and Harriet’s first child will have the same blood type as either of the parents?
Chance for A = 1/2 (chance of allele i from Harold) x [(2/3 x 1/2) + [1/3 x 1)] (chance of IA allele from Harriet)
= 1/2 x 2/3 = 1/3
Chance for B = 1/2 (chance of allele IB from Harold) x [(2/3 x 1/2) + (1/3 x 0)] (chance of i allele from Harriet)
= 1/2 x 1/3 = 1/6
Chance of either type A or type B = 1/3 + 1/6 =1/2
4
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
(4) Michael (15 pts) You are studying the genetics of pumpkin shape. You have purchased two plants from true-breeding
lines, one from a variety called ribbed and another from a variety called smooth. You cross these two pumpkin plants and
find that the F1 plants all produce pumpkins with a new phenotype that you label warty. You allow the warty F1 pumpkin
plants to self-fertilize and observe that the F2 generation of pumpkin plants includes 21 ribbed, 29 smooth, and 63 warty.
When you self-fertilize the warty F2 pumpkins, you find that 7 of the 63 are true-breeding.
a) (3 pts) How is pumpkin shape inherited? Be specific (ten word maximum).
Two genes with recessive epistasis determine shape.
b) (3 pts) What the genotypes of the parents, the F1, and the three classes of F2 plants.
(AAbb ribbed x aaBB smooth produced AaBb warty F1 plants. Self-fertilization of AaBb produced 9 A_B_ warty : 3 A_bb
ribbed : 4 aaB_ or aabb smooth.)
c) (3 pts) What results do you predict for the cross between the F1 warty pumpkin plant and the parental ribbed pumpkin
plant? On the lines, provide the expected fraction of pumpkin plants of each variety among the progeny.
Ribbed: 1/2
Smooth: 0
Warty: 1/2
AaBb x AAbb will produce 1/2 A_Bb warty and 1/2 A_bb ribbed.
5
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
d) (3 pts) What results do you predict for a cross between the F1 warty pumpkin plant and an F2 smooth pumpkin plant?
Assume that the F2 plant used in the cross has the genotype that is most common out of all the possible genotypes of F2
smooth plants. On the lines, provide the expected fraction of pumpkin plants of each variety among the progeny.
Ribbed: 1/8
Smooth: 1/2
Warty: 3/8
F2 smooth plants can be aaBB (1/4) aaBb (1/2), or aabb (1/4). The most common genotype for a smooth plant is aaBb.
AaBb x aaBb will produce 1/8 Aabb ribbed, 1/2 smooth (aa B_ or aabb), and 3/8 AaB_ warty
e) (3 pts) You continue to self-fertilize the F1 warty pumpkins to produce hundreds of F2 progeny. From the ribbed F2
progeny plants, you randomly pick two plants. You cross these two ribbed pumpkins together and then only plant one seed to
produce one progeny plant. What is the probability that this plant will have the smooth phenotype?
1/9
Ribbed F2 plants can be AAbb (1/3 chance) or Aabb (2/3 chance).
The progeny from this cross can be either ribbed (A_bb) or smooth (aabb).
P(smooth progeny) = P(aabb genotype) = P(Aabb x Aabb produce aabb progeny)
= 2/3 x 2/3 x 1/4 = 1/9
6
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
(5) Bella (12 pts) Duchenne-type muscular dystrophy is an inherited human disease, and one type of color blindness is sexlinked. Two healthy parents with normal color vision have many children. All of their daughters are healthy and have normal
vision. However, half their sons are healthy but are color blind; the other half have normal vision but have Duchenne disease.
a) (4 pts) Is Duchenne disease an autosomal or a sex-linked trait? Explain your answer.
It is sex linked, just like color blindness is.
b) (4 pts) Diagram the probable genotype of the parents
Female is xmd+x+cb and male is x++y
c) (4 pts) Could such parents ever have a son with neither Duchenne nor colorblindness? If so, how?
The couple could have an unaffected son if a recombination on the X chromosome between the Duchenne and colorblindness
mutations yielded an egg that did not carry either mutation.
7
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
(6) Manon (8 pts) Fruit color and plant height in a particular species of plant are traits that are each controlled by a single
gene. This plant has only 2 pairs of chromosomes and the gene for each trait is on a different chromosome pair. In a plant
that is heterozygous (AaBb), sketch the two chromosomes in each of the following phases. In your sketch, indicate the
location of the alleles. You must draw all chromatids of all homologous chromosomes.
a) (1 pt each) In mitosis (if multiple cells, draw all cells):
Metaphase
Anaphase
b) (2 pts each) In meiosis (if multiple cells, draw all cells):
Metaphase I
Anaphase I
8
Anaphase II
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
(7) Kevin (10 pts) A geneticist is studying nondisjunction in a small population of people with instances of a peculiar trait,
the ability to fly. He traces the ability to fly to a recessive allele on the X-chromosome. In a particular family, a normal
woman marries a normal man and together they give birth to a boy with Klinefelter syndrome (XXY) and the ability to fly.
The geneticist is able to conclude that nondisjunction occurred in only one of the parents.
a) (4 pts) In which parent did nondisjunction occur and at what stage of meiosis? Show your work.
Mother, meiosis II
b) (2 pts) The geneticist discovers that the people in this small population actually have 24 pairs of chromosomes (1 extra set
of autosomal chromosomes). How many homologous chromosomes would the boy described above have in one of his
somatic cells (for instance a skin cell)?
49
c) (4 pts) If nondisjunction were to occur in the father at meiosis I, list all of the possible genotypes and syndromes of their
male and female offspring. Recall that XO individuals have Turner syndrome while XXY individuals have Klinefelter
syndrome.
Xf: flying female (Turner syndrome)
Xf+: non flying female (Turner syndrome)
Xf+Xf+Y: non flying male (Klinefelter syndrome)
XfXf+Y: non flying male (Klinefelter syndrome)
9
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
(8) Ao (12 pts) Hobbits usually have hairy feet (H) and slightly pointed ears (E). Scientists found that genes responsible for
hairy feet and slightly pointed ears are on the same autosomal chromosome and they are 20 map units apart. Scientists also
discovered that the recessive alleles of these two genes give hobbit very pointed elfish ears (e) and hairless feet (h),
respectively.
An elfish ear female hobbit married a hairless foot male hobbit and they gave birth to a daughter with normal appearance and
a son with elfish ears and hairless feet.
a) (4 pts) Assign genotypes to the parents, to the daughter, and to the son.
Mother: eH/eh Father: Eh/eh
Daughter eH/Eh
Son eh/eh
b) (4 pts) What gametes and in what proportions will the daughter produce?
eH:Eh:eh:EH=0.4:0.4:0.1:0.1
c) (4 pts) If an individual with the same genotype as the daughter marries an individual with the same genotype as the son,
what is the chance that this couple would produce a child with elfish ears and hairless feet?
0.1
10
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
Thursday, October 31, 2013
STUDENT NAME: _______________________________________________
(Last)
(First)
ID # _________________
(9) Josh (15 pts) In spiders, the abdomen shape (determined by gene A) can be either spherical or elliptical, the color
(determined by gene C) can be orange or black, and the legs (determined by gene L) can be furry or hairless. These traits are
each determined by one gene with two alleles. You cross a true breeding spherical, orange, furry spider to a true breeding
elliptical, black, hairless spider, and the F1 progeny are all elliptical, orange, and hairless.
a) (3 pts) You want to perform a test cross with the F1 progeny. What type of spider should you use to mate with the F1
spider? List the phenotype for abdomen, color, and legs for the spider you will choose.
Spherical
black
furry
The F1 spider shows that elliptical, orange, and hairless are each dominant traits. The test cross should use a spider recessive
for all three traits.
b) (2 pts) If the these three genes assort independently, what fraction of the progeny from the test cross in part a) do you
expect to have the same phenotype as the F1 spider (elliptical, orange, hairless)?
1/8
The F1 spider is heterozygous for each gene. The probability of obtaining 3 dominant alleles is (1/2)^3.
When you perform the test cross of the F1 spider, you obtain the following 200 progeny:
63
25
3
10
8
5
29
57
spherical
spherical
spherical
spherical
elliptical
elliptical
elliptical
elliptical
orange
black
orange
black
orange
black
orange
black
furry
furry
hairless
hairless
furry
furry
hairless
hairless
c) (6 pts) What is the map unit distance for each of the following gene pairs?
A-C: 44
A-L: 13
C-L: 31
A-C: 13 + 31 = 44 map units
A-L: (8 + 10 + 3 + 5) / 200 = 26/200 = .13 = 13 map units
C-L: (25 + 29 + 3 + 5) / 200 = 62/200 = .31 = 31 map units
d) (4 pts) After performing the test cross described above, you now cross F1 siblings together. What percent of the progeny
from this sibling cross do you expect to be black and furry?
2.40%
The color and leg genes are 31 map units apart and the parental types are orange/furry and black/hairless.
The chance an F1 spider produces a gamete with the black & furry alleles is 15.5%.
The probability of getting the black and furry recessive alleles from each parent is (0.155)^2 = .00240 = 2.4%.
11
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
(10) Bella (14 pts) A cross was performed between an Hfr arg-met+lys+trp+ and a F- arg+met-lys-trp-. A master plate was
used to select for arg+ trp+ colonies. Finally, replica plating was used to screen for the other markers. The following results
were obtained:
Plate
Master
Min+met
Min+lys
Min
Number of colonies
500
20
335
15
Genotypes
trp+ arg+ met+/- lys+/trp+ arg+ met+/- lys+
trp+ arg+ met+ lys+/trp+ arg+ met+ lys+
a) (2 pts) What was added to the master plate in addition to the Minimal Medium?
MM+met+lys
b) (2 pts) How many colonies have the genotype met+ lys+ trp+?
15
c) (2 pts) How many colonies have the genotype met- lys+ trp+?
20-15=5
d) (2 pts) How many colonies have the genotype met+ lys- trp+?
335-15=320
e) (3 pts) How many colonies have the genotype met- lys- trp+?
500-15-5-320=160
f) (3 pts) What fraction of the colonies represent events that would include crossovers between trp and met genes?
It is all the colonies that are met – trp +.
(160+5)/500 = 33%
or
(500-335)/500 = 33%
12
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
Thursday, October 31, 2013
STUDENT NAME: _______________________________________________
(Last)
(First)
ID # _________________
(11) Linda (11 pts) In a generalized transduction experiment, you grew bacteriophage on plates containing two types of
bacteria. One of these strains is Ala+ Cys+ Glu+ Met+ Phe+ Tyr-. The other is Ala+ Cys+ Glu+ Met+ Phe+ Tyr+.
However, you mixed up the strains so do not know which strain is which. You use each strain to transduce Ala- Cys- GluMet- Phe- Tyr- bacteria. The table below shows the results of selecting on master plates containing all but the listed nutrient,
and replica plating on plates containing all but the listed nutrient.
Strain A: 800 colonies on each master plate
Replica
Replica
plate (-Ala) plate (-Cys)
Master plate (-Ala) x
145
Master plate (-Cys)
x
Master plate (-Glu)
Master plate (-Met)
Master plate (-Phe)
Strain B: 800 colonies on each master plate
Replica
Replica
plate (-Ala) plate (-Cys)
Master plate (-Ala) x
142
Master plate (-Cys)
x
Master plate (-Glu)
Master plate (-Met)
Master plate (-Phe)
Replica
plate (-Glu)
140
650
x
Replica
plate (-Met)
550
230
210
x
Replica
plate (-Phe)
480
320
350
600
x
Replica
plate (-Tyr)
0
5
7
0
0
Replica
plate (-Glu)
145
649
x
Replica
plate (-Met)
550
228
215
x
Replica
plate (-Phe)
481
324
354
599
x
Replica
plate (-Tyr)
0
0
0
0
0
a) (3 pts) Which strain is Tyr+? How do you know?
Strain A – it has colonies when plated on Tyr- plates, whereas Strain B has no colonies.
b) (3 pts) What is the co transduction frequency between Ala and Met?
550/800 = 68.75%
c) (5 pts) What is the order of the genes. Explain your answer, including any ambiguities.
Tyr Glu/Cys Phe Met Ala
13
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
Thursday, October 31, 2013
STUDENT NAME: _______________________________________________
(Last)
(First)
ID # _________________
(12) Josh (14 pts) You have a prototrophic Hfr strain that is resistant to erythromycin and unable to utilize the sugar mannose
and an F- strain with genotype ala- cys- eryS man+. You know that the cys and ery markers are transferred within 20 minutes
of conjugation, but ala and man markers are not transferred if conjugation is interrupted at 20 minutes. Also the ala gene can
be transferred if conjugation proceeds for 40 minutes, but the man gene is far away from the origin. You wish to set up a
three point cross map to map these genes. You will allow conjugation for 40 minutes with the goal of allowing at least three
markers the potential to be transferred from the donor to recipient.
a) (3 pts) What media should you use to select for exconjugants? Circle one answer for each sentence.
Carbon source(s) on the master plate should be:
glucose
The master plate should include:
The master plate should include:
alanine
cysteine
mannose
both
erythromycin
no antibiotic
both nutrients
neither nutrient
Answer: mannose; no antibiotic; cysteine.
You allow bacteria to undergo conjugation for 40 minutes, then spread the bacteria on the master plate. Your master plate has
500 colonies. You then make the following replica plates and observe number of colonies. (MM = minimal media with
glucose, unless mannose is added, in which case mannose is the only carbon source.)
Replica Plate
1
2
3
4
5
6
7
8
Media
MM + cys
MM + cys + man
MM + cys + ery
MM + cys + ery + man
MM + ery
MM + ery + man
MM
MM + man
Number of colonies on plate
500
500
230
230
140
140
155
155
b) (2 pts) Which plate(s) allow the growth of colonies with genotype cys- man+? Circle all correct answers.
1
2
3
4
5
6
7
Answer: Plates 1, 2, 3, 4
14
8
none of the 8 replica plates
INSTRUCTORS: Steve Jacobsen
Daniel Cohn
LS4 MIDTERM
STUDENT NAME: _______________________________________________
(Last)
(First)
Thursday, October 31, 2013
ID # _________________
c) (6 pts) Of the 500 colonies on the master plate, how many have each of the following genotypes?
cys+ eryR: 140
cys- eryS: 255
cys+ eryR: 140 (look at plate 5 or 6)
cys- eryR: 90 (plate 3 or plate 4 count – plate 5 count)
cys+ eryS: 15 (plate 7 or plate 8 count – plate 5 count)
cys- eryS: 500 – 140 – 15 – 90 = 255 colonies
d) (3 pts) Based on the colony counts on the replica plates, which gene is likely closest to the origin of transfer in your Hfr
strain? Explain your reasoning to receive credit.
Cys is likely closest to the origin. The genotype cys+ eryS is the rarest of the 4 possible genotypes on the master plate. If the
order is cys first then ery, this genotype would require a quadruple crossover. If the order was ery first then cys, then this
genotype would only require a double crossover, and then should not be the rarest genotype.
15