Quiz 6 Question Pool Spring 2022
1) What is the overall goal for the Molecular Biology lab? (2 points)
- To combine DNA from different sources into a single DNA molecule
2) Molecular biologists use a number of normal cellular enzymes to manipulate genes. Name 3 of them
& briefly describe their functions in a cell. (3 points)
- Lysosome - degrades peptidoglycans
- DNA ligase - closes gaps between Okazaki fragments during DNA replication
- Restriction enzymes - cut an internal site in a strand of DNA
3) a. Why must plasmids have selectable markers? (1 point per reason up to 2 points)
- Isolate the desired plasmid
- Plasmids can be lost each generation so selectable markers ensure that they are taken up by
the bacteria
b. What is the name of the plasmid that we’ll be using? (1 point)
- pUK21
c. What is the selectable marker in our plasmid? (1 point)
- Kanamycin resistance
4) Give two similarities between plasmids & bacterial chromosomes (besides the fact that they’re both
DNA). (2 points)
- Circular
- self replicating and heritable
5) a. What is the doubling period of E. coli under optimum conditions? (1 point)
- 20 minutes
b. If you start a culture at 5 PM with one E. coli cell, how many will you have at 8 AM the following
morning, assuming that the bacteria experience optimum conditions all night? Show your work! (2
points)
- 15hrs x 3doubling periods/hour = 45 doubling periods
- 2^45 bacteria
c. In fact, bacteria use up all the nutrients in standard medium by the time they reach a density of about
1010 cell/ml. In order to feed all the bacteria in part (b), how many liters of medium would you need? To
put this in perspective, remember 1 liter is about 1 quart! (2 points)
6) a. Describe restriction enzymes: what are they & what do they do? Include some detail! (3 points)
- They are enzymes that cut an internal site in a strand of DNA
- Usually cut a palindromic sequence of 4-8 nucleotides
- Some restriction enzymes create blunt ends while other result in sticky ends
b. Why do bacteria make restriction enzymes? (2 points)
- Defend against bacteriophages
- digest phage’s genome when it enters the bacterial cell
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7) Draw a picture which summarizes ligating a gene into a plasmid. (3 points)
8) a. What are 2 names for the region of a plasmid into which foreign DNA is usually inserted? (2
points)
- Vector or Backbone
b. Briefly describe this region. What does it consist of? (2 points)
- An origin of replication and a selectable marker
9) What are 2 strategies to prevent a linearized vector from closing on itself during a ligation? (2 points)
- add 3x the amount of gene fragments as vectors
- cut the gene fragment and vector with two incompatible restriction enzymes
10) What is transformation? (2 points)
- When bacteria take up foreign DNA
11) What are competent cells? (2 points)
- cells with altered cell walls so that they can take up DNA
12) a. In order to do a blue/white screen, where must the 2 parts of the lacZ (β-galactosidase) gene be?
(2 points)
- Larger part - E. coli chromosome
- Smaller part - cloning vector
b. If both parts are present & functional inside an E. coli cell, what color will a colony grown from
that cell on X-gal be? (1 point)
- Blue
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Quiz 5 Question Pool Spring 2022
ALWAYS SHOW YOUR WORK! Avoid the use of ambiguous symbols for dominant vs. recessive alleles.
Problem 4-17:
In some plants a red pigment, cyanidin, is synthesized from a colorless precursor. The addition of a hydroxyl group (OH-) to the
cyanidin molecule causes it to become purple. In a cross between two randomly selected purple plants, the following results were
obtained:
94 purple
31 red
43 white
a. How many genes are involved in the determination of these flower colors? (1 point)
Dihybrid ratio of 9:3:4 which means two genes are involved.
b. Which genotypic combinations produce which phenotypes? Diagram the purple x purple cross. (4 points)
Purple - AaBb
red - Aabb
colorless - aabb
Problem 2-5 (3-5 in 6th edition):
In Drosophila melanogaster, very dark (ebony) body color is determined by the e allele. The e allele produces the normal
wild-type honey-colored body. In heterozygotes for the two alleles, a dark marking called the trident can be seen on the thorax,
but otherwise the body is honey-colored. The e allele is thus considered to be incompletely dominant to the e allele.
a. When female e e flies are crossed to male e e flies, what is the probability that the progeny will have the dark trident marking?
(1 point)
½ the flies will have dark markings
b. Animals with the trident marking mate among themselves. Of 300 progeny, how many would be expected to have a trident,
how many ebony bodies, and how many honey-colored bodies? (2 points)
¼ or 75 will have honey marking, ½ or 150 will have trident marking, and ¼ or 75 will have ebony markings
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+
+
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Problem 2-8:
A wild legume with white flowers and long pods is crossed to one with purple flowers and short pods. The F offspring are
allowed to self-fertilize, and the F generation has 301 long purple, 99 short purple, 612 long pink, 195 short pink, 295 long
white, and 98 short white progeny.
a. How are these traits being inherited? (2 points)
Flower color caused by incompletely dominant alleles of a gene, w C^p giving purple when homozygous, C^w giving white
when homozygous, ad the C^p C^w heterozygous giving pink. Pod shape is controlled by a single gene w the long allele
completely dominant to the short allele
b. What did the F generation look like? (1 point)
White long x purple short → 301 long purple: 99 short purple: 612 long pink: 195 short pink: 295 long white: 98 short white
1
2
1
Problem 2-10 (3-10 in 6th edition):
There are several genes in humans in addition to the ABO gene that give rise to recognizable antigens on the surface of red blood
cells. The MN and Rh genes are two examples. The Rh locus can contain either a positive or negative allele, with positive being
dominant to negative. M and N are codominant alleles of the MN gene. The following chart shows several mothers and their
children. For each mother-child pair, choose the father of the child from among the males in the right column, assuming one child
per male. (4 points)
a.
b.
c.
d.
Mother
O M Rh pos
B MN Rh neg
O M Rh pos
AB N Rh neg
Child
B MN Rh neg
O N Rh neg
A M Rh neg
B MN Rh neg
Males
(1) O M Rh neg
(2) A M Rh pos
(3) O MN Rh pos
(4) B MN Rh pos
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Father of (a) is number 4
Father of (b) is number 3
Father of (c) is number 2
Father of (d) is number 1
Problem 2-15:
Fruit flies with one allele for curly wings (Cy) and one allele for normal wings (Cy ) have curly wings. When two curly-winged
flies were crossed, 203 curly-winged and 98 normal-winged flies were obtained. In fact, all crosses between curly-winged flies
produce nearly the same curly:normal ratio among the progeny.
a. What is the approximate phenotypic ratio in these offspring? (1 point)
⅔ Curly: ⅓ normal
b. Suggest an explanation for these data. (2 points)
The expected result for this cross is: Cy+Cy x Cy+Cy → ¼ CyCy (?) : ½ Cy+Cy (curly): ¼ Cy+Cy+ (normal). If the CyCy
genotype is lethal, then the expected ratio will match the observed data.
c. If a curly-winged fly was mated to a normal-winged fly, how many flies of each type would you expect among 180 total
offspring? (2 points)
The cross is Cy+Cy x Cy+Cy+ → ½ Cy+Cy: ½ Cy+Cy+, so there would be approximately 90 curly-winged and 90
normal-winged flies
+
Problem 2-22 (3-20 in 6th edition):
A rooster with a particular comb morphology called walnut was crossed to a hen with a type of comb morphology known as
single. The F progeny all had walnut combs. When F males and females were crossed to each other, 93 walnut and 11 single
combs were seen among the F progeny, but there were also 29 birds with a new kind of comb called rose and 32 birds with
another new comb type called pea.
1
1
2
a. Explain how comb morphology is inherited. (3 points)
Four F2 phenotype means there are two genes, A and B. Both genes affect the same structure i.e. comb. The F2 phenotypic
dihybrid ratio among the progeny is 9:3:3:1. Since walnut is the most abundant F2 phenotype, it must be due to A-B genotype.
Single combs are the least frequent class, and are thus aabb.
If the walnut F2 is A-B, then the original walnut parent must have been AABB,and if AABB X aabb, gives AaBb (walnut) i.e.
9/16 A-B (walnut) : 3/16 A-bb (rose): 3/16 aa B-(pea): 1/16 aabb (single)
b. What progeny would result from crossing a homozygous rose-combed hen with a homozygous pea-combed rooster? What
phenotypes and ratios would be seen in the F2 progeny? (3 points)
If homozygous rose-combed hen crossed with a homozygous pea-combed rooster, then F1 progeny result in walnut comb i.e.
AaBb which in F2 generation gives walnut 9: rose 3: pea 3: single
c. A particular walnut rooster was crossed to a pea hen, and the progeny consisted of 12 walnut, 11 pea, 3 rose, and 4 single
chickens. What are the likely genotypes of the parents? (2 points)
Genotypes of parents are AaBb and aaBb
Problem 2-23:
A black mare crossed to a chestnut stallion produced a bay son and a bay daughter. The two offspring were mated to each other
several times, and they produced offspring of four different coat colors: black, bay, chestnut, and liver. Crossing a liver grandson
back to the black mare gave a black foal, and crossing a liver granddaughter back to the chestnut stallion gave a chestnut foal.
Explain how coat color is being inherited in these horses. (3 points)
4 genotypes in the F2 generation means 2 genes determine coat color. The F1 bay animals produce 4 phenotypic classes, so they
must be dihybrids, Aa Bb. The liver horses alleles don't affect color, suggesting the recessive genotype aa bb. Although its
probable that the original black mare was AA bb and the chestnut stallion was aa BB, each of these animals produced only 3
progeny, so it cant be concluded definitely that these animals were homozygous for the dominant allele they carry. Therefore, the
black mare A- bb, the chestnut stallion was aa B-, and the F1 bay animals are Aa Bb. The F2 horses were: bay (A- B-), liver (aa
bb), chesnut (aa B-), and black (A- bb).
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Problem 2-29 (3-24 in 6th edition):
Two true-breeding white strains of the plant Illegitimi noncarborundum were mated, and the F progeny were all white. When the
F plants were allowed to self-fertilize, 126 white-flowered and 33 purple-flowered F plants grew.
a. How could you describe the inheritance of flower color? Describe how the specific alleles influence each other and therefore
affect phenotype. (3 points)
The cross is due to two genes. The F2 phenotypic ratio is thus approximately 13 white :3 purple. The data fit the hypothesis that
two genes control color, and that the F1 are dihybrids.
AA BB (white) × aa bb (white) → Aa Bb (white, same as AA BB parent) →9 A– B– (white) : 3 A– bb (unknown) : 3 aa B–
(unknown) : 1 aa bb (white).
1
1
2
b. A white F plant is allowed to self-fertilize. Of the progeny, ¾ are white-flowered and ¼ are purple-flowered. What is the
genotype of the white F plant? (1 point)
Aa BB (white)
2
2
c. A purple F plant is allowed to self-fertilize. Of the progeny, ¾ are purple-flowered and ¼ are white-flowered. What is the
genotype of the purple F plant? (1 point)
aa Bb.
2
2
Problem 2-26:
Filled-in symbols in the pedigree that follows designate individuals who are deaf.
a. Study the pedigree and explain how deafness is being inherited. (2 points)
Bc unaffected individuals had affected children, deafness in this pedigree is caused by homozygosity for a recessive allele. From
affected individual II-1, you know I-1 and I-2 are carriers. The trait was passed on to generation III thru II-2 who was also a
carrier. All children of affected individuals III-2 x III-3 are affected, as predicted for a recessive trait. However, generation V
seems inconsistent w inheritance of a recessive allele of a single gene. This result is consistent w 2 diff genes involved in hearing
w a defect in either gene leading to deafness: the trait is heterogeneous, meaning the 2 family lines shown are homozygous for
recessive mutant (deafness) alleles of 2 separate genes.
b. What is the genotype of the individuals in generation V? Why are they not deaf? (2 points)
Individuals in V are doubly heterozygous (Aa Bb), having inherited a dominant & recessive allele of each gene from their
parents (aa BB x AA bb). The ppl in V are unaffected bc one dominant allele of each gene is sufficient for normal function.
Problem 2-38:
The Bombay phenotype is important not only in solving paternity cases, but it also has important implications for blood
transfusions involving people with this condition. The issue is that individuals can make antibodies against foreign blood cell
molecules not in their own bodies. From which donors can Bombay individuals safely receive transfusions, and to which
recipients can they donate their own blood? Explain your reasoning. (4 points)
Bombay individuals can accept a transfusion of only blood cells w/out H sugars on their surfaces–only blood cells from other
Bombay individuals. However, ppl, w the Bombay phenotype are universal donors
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Problem 2-39 (3-32):
Three different pure-breeding strains of corn that produce ears with white kernels were crossed to each other. In each case, the F
plants were all red, while both red and white kernels were observed in the F generation in a 9:7 ratio. These results are tabulated
here.
F
F
white-1 x white-2
red
9 red: 7 white
white-1 x white-3
red
9 red: 7 white
white-2 x white-3
red
9 red: 7 white
1
2
1
2
a. How many genes are involved in determining kernel color in these three strains? (1 point)
3 genes
b. Define your symbols and show the genotypes for the pure-breeding strains white-1, white-2, and white-3. (1 point)
White-1 is aa BB CC;
white-2 is AA bb CC;
white-3 is AA BB cc .
c. Diagram the cross between white-1 and white-2, showing the genotypes and phenotypes of the F and F progeny. Explain the
observed 9:7 ratio. (3 points)
aa BB CC (white-1) × AA bb CC (white-2) →
Aa Bb CC (red) →9/16
A– B– CC (red) : 3/16
A– bb CC (white) : 3/16
aa B– CC (white) :
1/16 aa bb CC (white).
Red color requires a dominant, functional allele of each of the three genes (A– B– C–).
1
2
Problem 2-41 (3-33 in 6th edition):
In mice, the A allele of the agouti gene is a recessive lethal allele, but it is dominant for yellow coat color. What phenotypes and
ratios of offspring would you expect from the cross of a mouse heterozygous at the agouti gene (genotype A A) and also at the
albino gene (Cc) to an albino mouse (cc) heterozygous at the agouti gene (A A)? (3 points)
2/6 AyA Cc (yellow): 3/6 - - cc (albino): ⅙ AA Cc (agouti)
Y
Y
Y
Problem 2-46:
You picked up two mice (one female and one male) that had escaped from experimental cages in the animal facility. One mouse
is yellow in color, and the other is brown agouti. (Agouti hairs have bands of yellow, while non-agouti hairs are solid-colored).
You know that this mouse colony has animals with different alleles at only three coat color genes: the agouti (A) or non-agouti (a)
or yellow (A ) alleles of the A gene (A > A > a; A is a recessive lethal), the black (B) or brown (b) alleles of the B gene (B > b),
and the albino (c) or non-albino (C) alleles of the C gene (C > c; cc is epistatic to all alleles of the other two genes). However,
Y
Y
Y
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you do not know which of the alleles are actually present in each of the animals you’ve captured. To determine the genotypes,
you breed the two escaped mice together. The first litter has only three pups. One is albino, one is brown (non-agouti), and the
third is black agouti.
a. What alleles of the A, B, and C genes are present in the two mice you caught? (2 points)
Ay a Bb Cc × Aa bb Cc.
b. After raising several litters from these two parents, you have many offspring. What phenotypes in what ratios do you expect to
see? (4 points)
six different coat colors are possible: albino (–– –– cc), yellow [Ay (A or a) –– C–], brown agouti (A– bb C–),
black agouti (A– B– C–), brown (aa bb C–), and black (aa B– C–).
Problem 2-49 (3-41 in 6th edition):
The garden flower Salpiglossis sinuate (“painted tongue”) comes in many different colors. Several crosses are made between
true-breeding parental strains to produce F plants, which are in turn self-fertilized to produce F progeny.
Parents
F phenotypes
F phenotypes
red X blue
all red
102 red, 33 blue
lavender X blue
all lavender
149 lavender, 51 blue
lavender X red
all bronze
84 bronze, 43 red, 41 lavender
red X yellow
all red
133 red, 58 yellow, 43 blue
yellow X blue
all lavender
183 lavender, 81 yellow, 59 blue
1
1
2
2
State a hypothesis explaining the inheritance of flower color in painted tongues and assign genotypes to the parents, F progeny,
and F progeny for all 5 crosses. (5 points)
For 3 points extra credit, draw a pathway which illustrates your hypothesis!
1. 1 gene red or blue
2. 1 gene lavender or blue
3. One gene, codominance/incomplete dominance (1:2:1), the heterozygote is bronze.
4. Two genes with recessive epistasis (9 red : 4 yellow : 3 blue).
5. Two genes with recessive epistasis (9 lavender : 4 yellow : 3 blue).
In total there are two genes. One gene determines blue (cb ), red (Cr ) and lavender (Cl ) where Cr = Cl > cb. A second gene
controls yellow: Y seems to have no effect on color, so in the presence of Y the color is determined by the alleles of the C gene.
The y allele makes the flower yellow, and yy is epistatic to all alleles of the C gene.
1
2
Problem 2-53:
A couple wants to know the probability that their expected child will suffer from split-hand deformity, which affects the
prospective father, who is indicated by an arrow in the pedigree shown. (The arrow means that he is the proband - the person in
the family who first brought the disorder to the attention of medical professionals). This trait is rare in the population, and the
prospective parents are not related to each other.
a. What is the mode of inheritance of this trait? (2 points)
Dominant
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b. What is the penetrance of this trait [that is, the ratio between the number of individuals in the pedigree who display the trait
(numerator), and the number of individuals you know from the pedigree must have the trait-determining genotype regardless of
whether they have the trait or not (denominator)]? (2 points)
At most ⅚ = 83%
c. Using your answer to part (b), what would you tell the parents about the numerical likelihood their expected child will have
split-hand deformity? (2 points)
½ x 83% = 42%
d. Why is it possible that the likelihood the child will be affected is actually less than the number you just answered in part (c)?
You should specify the lowest numerical likelihood that could possibly be consistent with the data. (3 points)
4 ppl in the pedigree (III-1, III-2, III-4, and III-7) might have the mutant allele (bc their parents had the mutant allele), but no info
in the pedigree allows us to know one way or the other.
Problem 2-55 (3-47 in 6th edition):
Spherocytosis is an inherited blood disease in which the erythrocytes (red blood cells) are spherical instead of biconcave. This
condition is inherited in a dominant fashion, with ANK1 (the nonfunctional mutant allele) dominant to ANK1 . In people with
spherocytosis, the spleen recognizes the spherical red blood cells as defective and removes them from the bloodstream, leading to
anemia. The spleen in different people removes the spherical erythrocytes with different efficiencies. Some people with spherical
erythrocytes suffer severe anemia and some mild anemia; yet others have spleens that function so poorly there are no symptoms
of anemia at all. When 2400 people with the genotype ANK1/ ANK1 were examined, it was found that all of them had spherical
erythrocytes, 2250 had anemia of varying severity, and 150 had no anemia symptoms. Does this description of people with
spherocytosis represent incomplete penetrance, variable expressivity, or both? Explain your answer. Can you derive any values
from the numerical data to measure penetrance or expressivity? (3 points)
All ppl w the genotype ANK1+ ANK1 have spherical erythrocytes instead of normally concave ones. Therefore, the spherical
character is fully penetrant & shows no variation in expression. The second trait is anemia. The expressivity among anemic
patients varies from severe to mild. In fact, some ppl w the ANK1+ ANK1 genotype (150/2400) have no symptoms of anemia at
all. Thus, the penetrance of anemia is 2250/2400 or 0.94.
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