Solutions to 7.014 Problem Set 7
Question 1
The plasmid vector pet83 contains an ampicillin resistance gene (Amp) and a kanamycin
resistance gene (Kan). The plasmid is cut at the BamHI site shown below, and successfuly
ligated with a compatible human BamHI fragment. The recombinant plasmid is then
transformed into E. Coli. The transformation mixture is then spread on petri plates and
allowed to grow.
a) What type of E. Coli cells would you choose as the host strain for your transformation?
You would begin with a strain that is ampSkan.
b) What antibiotic would you include in the medium to ensure that only bacteria containing
the pet83 plasmid will grow?
Ampicillin
c) What two antibiotic resistance phenotypes will be found among the transformed bacteria?
ampRkanR and ampRkanS
d) Of the antibiotic resistance phenotypes listed above, which will contain the human BamHI
fragment?
ampRkanS
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e) The human BamHI fragment that you cloned into pet83 encodes the enzyme helicase. You
decide to clone the mouse version of the helicase gene.
i) Briefly explain how you would make a mouse genomic library in E. Coli.
To make a mouse genomic library in bacteria:
1) Digest mouse genomic DNA with a restriction enzyme and digest a plasmid vector with the
same restriction enzyme.
2) Ligate the mouse DNA into the vector.
3) Transform ampicillin and kanamycin sensitive bacteria with this ligation mixture.
4) Plate the transformed bacteria on media containing ampicillin. This selects for cells that are
carrying a plasmid.
ii) Once you have the mouse genomic library in E. Coli, how would you find the cells
that contain the mouse helicase gene?
When thinking about this question, recognize that a mouse gene of similar structure and
function as a human gene will probably have a similar DNA sequence. We can take advantage of
this by assuming that DNA molecules with similar sequences will hybridize to each other.
1) Plate the E. Coli cells that compose the library on petri plates with ampicillin containing
media and make a copy or replica of the colonies on each plate using a piece of solid matrix
(nitrocellulose).
2) Lyse (break open) the bacteria on the nitrocellulose in such a way that single-stranded plasmid
DNA is accessible.
3) Wash the nitrocellulose with a solution containing radioactively labeled human helicase gene
as a hybridization probe to identify colonies containing DNA similar to this gene.
f) The restriction endonuclease sites for several enzymes are given below.
BamHI:
BglII:
EcoRI
t
t
5'...G G A T C C...3'
3'...C C T A G G...5'
t
5'...A G A T C T...3'
3'...T C T A G A...5'
s
PvuII
s
PvuI
s
s
EcoRV
t
5'...C A G C T G...3'
3'...G T C G A C...5'
5'...G A A T T C...3'
3'...C T T A A G...5'
t
5'...C G A T C G...3'
3'...G C T A G C...5'
s
t
5'...G A T A T C...3'
3'...C T A T A G...5'
s
i) DNA that has been cut with PvuII can be ligated to DNA that has been cut with
EcoRV. Indicate what would be the DNA sequence of the ligated DNA by filling in the
blanks below.
5'...C A G A T C...3'
3'...G T C T A G...5'
ii) EcoRI fragments can be ligated together because the ends have compatible
overhangs.
• Which two enzymes above create compatible ends? BamHI and BglII
•
If a DNA fragment cut with one of these enzymes was ligated to DNA cut with the
other, what would be the DNA sequence of the ligated DNA? Label 5' and 3' ends.
5'...G G A T C T...3'
3'...C C T A G A...5'
•
Can this new DNA be cut with either of the enzymes originally used? No.
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Question 1, continued
g) You successfully clone a DNA fragment (2.1 kb in length) carrying the mouse helicase gene
into a unique EcoRI restriction site of the plasmid pQE1 (4.1 kb in length). You digest this new
plasmid with the restriction enzymes EcoRI, BamHI, and PstI alone and in pairs. After running
out the digests on an agarose gel you get the following results. Lengths are in kb.
EcoRI
BamHI
PstI
EcoRI &
BamHI
BamHI
& PstI
EcoRI
& PstI
2.5
2.4
2.6
5.5
4.1
3.9
3.7
2.5
1.5
1.1
1.1
1.0
1.0
0.7
0.6
0.5
1.0
0.7
0.6
0.2
Draw a restriction map of the plasmid; include the fragment lengths, restriction sites, and label
the vector and the insert.
Note that the PstI and EcoRI site within
the insert are two separate sites. They
are close enough, however, to look like
one site in the restriction digest.
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Question 2
a) Complete the following life table for the gray squirrel.
x
0-1
1-2
2-3
3-4
4-5
5-6
6-7
7-8
8-9
nx
lx
530
134
56
39
23
12
5
2
0
1.00
0.253
0.106
0.074
0.043
0.023
0.009
0.004
0
Gray Squirrel Life Table
dx
mx
Lx
0.747
0.147
0.032
0.030
0.021
0.013
0.006
0.004
0
0.747
0.581
0.302
0.405
0.488
0.565
0.667
1.0
0
332
95
47.5
31
17.5
8.5
3.5
1.0
0
Tx
ex
536
204
109
61.5
30.5
13
4.5
1.0
0
1.01
1.52
1.95
1.58
1.33
1.08
0.90
0.50
0
b) Complete the Fecundity table for the gray squirrel.
x
0-1
1-2
2-3
3-4
4-5
5-6
6-7
7-8
8-9
lx
bx
lx b x
1.00
0.253
0.106
0.074
0.043
0.023
0.009
0.004
0
0.0
2.56
4.56
6.48
6.48
4.96
4.56
4.56
0
0.00
0.648
0.483
0.480
0.279
0.114
0.041
0.018
0
c) Is the population of squirrels increasing, decreasing, or remaining constant?
Ro = 2.06. Therefore the population is increasing.
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Question 2
a) Draw three different types of survivorship curves (label your axes) and briefly explain the
survival strategy associated with each.
100%
Type I
Type II
50%
Type III
0%
Age of Individuals in Cohort
Type I: Low mortality throughout life until an age threshold is reached, after which time individuals die at a relatively high
rate. This survival strategy usually entails parent(s) investing a lot of energy in raising their offspring, which keeps
mortality among young individuals low.
Type II: Constant death rate throughout life. Intermediate between types I and III, this survival strategy entails a moderate
input of energy from parent(s) in caring for their young. Populations which are subject to a constant rate of predation may
exhibit type II survivorship curves.
Type III: High mortality among the young; individuals which survive past young age tend to live for long time. This
strategy usually entails parents generating a lot of offspring without investing much energy in caring for any one individual.
The strategy is successful when at least a few of the many offspring survive their youth and reproduce.
b) For each of the following cohorts (group of individuals born at the same time), decide
which type of survivorship curve best applies. Explain briefly.
i) Cohort of 20 sugar maple seedlings growing in the shadow of the parent, a tall sugar
maple in mature forest.
Type III curve best applies, since the parent has generated many offspring but will invest no
extra energy in caring for these offspring. In fact, in this case the parent is successfully out
competing its offspring for the limited resource, light. When the tall sugar maple eventually
dies, one or two of its offspring may grow tall to fill the gap created in the canopy. The other
seedlings will remain small and will presumably die at a young age.
ii) Cohort of 100 babies born at Mass General Hospital.
Type I curve best applies, since human infant mortality is relatively low (when high quality
health care is accessible) and since humans invest a tremendous amount of energy in caring for
their offspring.
iii) Cohort of 1000 zooplankton in a lake containing a stable fish population which
preys on zooplankton.
Type II curve best applies. Parental zooplankton don't invest energy in caring for their
offspring. In this case, survivorship will mainly be determined by the rate at which the fish preys
on the zooplankton. Presumably zooplankton don't acquire skill at avoiding predation as they
age, so each zooplankton should be at an equal risk of predation throughout its life.
iv) Cohort of 100 corn plants planted and cultivated on a farm.
Type I. Plants will be protected from disease and predation and will live until they are harvested.
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Question 3
A breeding pair of rabbits is introduced to a secluded island off the coast of Australia. At first,
the population growth is far below K so the population may grow exponentially. The birth
rate of the rabbits is 4 rabbits/(rabbit•month) and the death rate is 0.5 rabbits/(rabbit•month).
a) What is the intrinsic growth rate of the rabbits?
rmax=intrinsic growth rate = birth rate - death rate = 4 – 0.5
rmax = 3.5 rabbits/ (rabbit month)
b) If the population grew exponentially, how many rabbits would there be after 15 months?
Ntotal = Noert
No = 2
r = intrinsic growth rate = 3.5 rabbits/(rabbit month)
t = time = 15 months
Ntotal = number of rabbits on the island at time t =1.26 x1023 rabbits
c) What is the doubling time (td) of this exponentially growing population?
2 No = Noe(rmaxTd) or 2 = or ln(2) = rTd
Therefore: doubling time = ln(2)/rmax = 0.197 months =~5.91 days
d) You observe the rabbit population over 15 months. You find that the specific rate of
increase is 1.2 rabbits/(rabbit•month). You tag the rabbits and find there are approximately
678 rabbits on the island. Using the density-dependent model of logistic growth…
i) calculate the carrying capacity of the island.
Specific rate of growth = (1/N)*(dN/dt) =1.2 rabbits/(rabbit month)
N = 678 rabbits
(1/N)*( dN/dt) = rmax((K-N)/K)
1.2/3.5 = ((K-678)/K)
K = 1031
ii) calculate the doubling time of this logistically growing population.
Doubling time = ln(2)/actual growth rate = ln(2)/1.2 = 0.577 month = ~17.3 days
e) What other factors besides density may play a role in controlling the actual growth rate of
the rabbit population on the island?
Some of the possibilities are competitors, predators, limiting nutrients (plant and water), and suitable
places to build homes.
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Question 4
Part1
You are studying farm snakes in Western Massachusetts. Farm snakes may be one of two
different colors, blue (BB or Bb) or red (bb).
genotype
BB
Bb
bb
# snakes
15
60
25
a) What is the frequency of the B allele (p)?
Total # of animals = 100, so total #o fallels = 200
P = # of B alleles/200 = 2 (15) + 60 /200 = 0.45
b) What is the frequency of the b allele (q)?
Total # of animals = 100, so total #o fallels = 200
P = # of b alleles/200 = 2 (25) + 60 /200 = 0.55
c) Is this population at Hardy-Weinberg equilibrium? Why or why not?
No, see below.
d) Given the p and q calculated above, how many snakes of each genotype are expected if the
population is in Hardy-Weinberg equilibrium?
genotype
BB
Bb
bb
# snakes
2
(.45) = .20
.20 x 100 =
(.45) (.55) (2) = .50
.50 x 100 =
(.55)2 = .30
.30 x 100 =
20
50
30
e) Why are real populations in nature not usually found at Hardy-Weinberg equilibrium?
The Hardy-Weinberg equation assumes that all of the following are true:
1. Infinite population size
2. Random mating occurs
3. No selective pressures are present
4. No mutations
5. No migration into or out of a population
In most populations, many of these conditions is broken.
f) Can a population be in Hardy-Weinberg equilibrium for one trait and not for another?
Yes.
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Question 4, continued
Part 2
Researchers isolated the XYZ protein from 12 different organisms (some extinct and some
present now) and determined the amino acid sequences of a particular region from each. The
sequences are listed below:
Organism
Sequence of XYZ
1
KSTSTDIKSREV
KSTSTDIKSREV
KTTATDIKSKEV
KSTAVEIKSKLF
KTTATEIKSKLV
KTLATDIKSREV
KSTATDIKSREV
KSTATEIKSKLV
KTTATEIKSKEV
KSTATDVKSREV
KSTAVEIKSKLV
KTTATDIKSREV
2
3
4
5
6
7
8
9
10
11
12
a) Construct one evolutionary tree that is consistent with this data. Indicate what
assumption(s) you have made.
Each of these sequences differs from its progenitor by a single amino acid substitution.
The sequence, #3, with the most different sequence from the others will be the common ancestor at the
base of the tree.
b) Which amino acid(s) are invariant throughout this evolutionary tree? Why might have
these amino acids been conserved in an actual organism?
The conserved sequence is K - - - - - - KS - - If these amino acids were important for maintaining the structure or the function of the protein in which
they are found, they might be conserved in evolution.
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