problem specimen assembled + answers
TRANSCRIPT
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Biology problem paper specimen
Please note that this is a specimen paper and you have been given eightspecimen questions to work through. This version contains thequestions with specimen answers.
One the actual paper there will be four questions, one each for Ecology,
Genetics, Cell and Organismal Biology and Molecular biology andbiochemistry.
The examination paper will be 3 hours in duration and you will beexpected to answer two questions from the four offered.
Each question will have a short summary at the beginning of the paper.
ECOLOGY 1This problem concerns root growth in grasses. How it can be measured, how it differs
between species and how it is affected by the presence and patchiness of nutrients and
organisms in the soil.
ECOLOGY 2This question requires you to analyse data from field surveys of butterflies in
grassland and woodland sites. You examine whether different types of butterfly
species differ in their abundance between habitats. You are asked to critically evaluate
the methods used by the student in their surveys, and to suggest new field work.
GENETICS 1This question is about genetic mapping in bacteria (which has been incredibly
important in the development of molecular biology) using conjugation, in which the
circular chromosome of the donor cell is broken at a fixed point and transferred as a
linear piece of DNA to a recipient.
GENETICS 2This problem asks if the difference between alleles of a mutant locus is significant and
if so, what might be the molecular explanation. Mutational changes in the DNA
sequences another mutant locus are shown and molecular explanations for their effect
are required.
CELL AND ORGANISMAL BIOLOGY 1Investigating the characteristics of the eWnt secreted signaling molecule. Analysis
and manipulation of DNA plasmid constructs coding for eWnt protein. Analyzing the
properties the eWnt protein and the eWnt gene promoter in vitro and in vivo.
CELL AND ORGANISMAL BIOLOGY 2This question concerns the use of oxygen by diving mammals. The first part of the
question is to investigate the relation between metabolic rate and size, and how this
may be influenced by their diving habits. The second part of the question concerns the
use of oxygen by human divers.
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MOLECULAR BIOLOGY AND BIOCHEMISTRY 1Loss of function mutations were generated on plasmids and investigated using yeast
cultures. Measurements were made so that the effect of the mutations could be
determined and hypotheses devised to explain the observations.
MOLECULAR BIOLOGY AND BIOCHEMISTRY 2Overexpression of protein at different temperatures: this problem concerns the
synthesis of a specific protein in a bacterial expression system. It deals with analysis
of bacterial growth, and characterisation of the product by gel chromatography.
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ECOLOGY 1
This problem concerns root growth in grasses. How it can be measured, how it
differs between species and how it is affected by the presence and patchiness of
nutrients and organisms in the soil.
STUDY 1
An ecology student interested in how five different grass species produced roots in an
organic nutrient patch (milled Lolium perenne shoot material) added to soil grew
plants from seed in microcosm units (dimensions 150 mm x 400 mm x 3 mm see
Figure). The organic material was added as a 20 mm band across the microcosm 200
mm below the soil surface. The remainder of the microcosm was filled with soil.
Seeds were planted 5 mm below the soil surface (see Figure).
Microcosms were placed in a growth cabinet under constant conditions of light,
humidity and water. After 42 days the student harvested the plants. Roots from the
organic patch were separated from the rest and their root length measured after which
these roots were dried at 70oC to obtain their dry weight (D.W.). Root length and dry
weights of the remaining roots were then recorded. These values were added to those
from the organic patch band to obtain the length and dry weight of the entire root
system (Table 1).
Table 1. Root length (m) and root D.W. (mg) for the five grass species in the organic
patch band and for the total root system.
Plant Species Root length
in organic
patch band
(m)
Total root
length
(m)
Root D.W. in
organic patch
band (mg)
Total root
D.W.
(mg)
Lolium perenne 3.569 12.0 22.0 2030
Festuca arundinacea 2.528 15.3 15.3 1870
Poa pratensis 1.707 12.1 8.1 1030
Dactylis glomerata 3.451 12.7 12.2 1670Phleum pratense 3.125 15.0 13.0 1720
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The student wished to calculate the root length density (RLD) which is the length of
roots per unit volume of soil (cm cm-3
) and specific root length (SRL) which is the
root length per unit root dry weight (cm mg-1
).
For each of the five plant species calculate:
a) The root length density (RLD; cm cm-3
) in the organic patch band. (3 marks)b) The specific root length (SRL; cm mg
-1) in the organic patch band. (4 marks)
Please show all your calculations.
Answers:
1. i) RLD in patch zone
Volume Organic Patch occupies = 0.3 x 2 x 15 = 9 cm3
Thus RLD = RL in Table 1 /100 to get cm then / 9 cm3
for RLD in patch
L. perenne = 356.9/9 = 39.7 cm cm-3
F. arundinacea = 252.8/9 = 28.1 cm cm-3
P. pratensis = 170.7/9 = 19.0 cm cm-3
D. glomerata = 345.1/9 = 38.3 cm cm-3
P. pratense = 312.5/9 = 34.7 cm cm-3
(3 marks)
1. ii) SRL (Nb: Convert RL to cm)
L. perenne = 356.9/22.0 = 16.2 cm mg-1
F. arundinacea = 252.8/15.3 = 16.5 cm mg-1
P. pratensis = 170.7/8.1 = 21.1 cm mg-1
D. glomerata = 345.1/12.2 = 28.3 cm mg-1
P. pratense = 312.5/13.0 = 24.0 cm mg-1
(4 marks)
STUDY 2
Using the same microcosm units the student then investigated how Lolium perenne
plants captured nitrogen (N) from a range of organic substrates of varying
carbon:nitrogen (C:N) ratio (Note: only one organic substrate was added to each
microcosm). After 55 days the L. perenne plants were removed from the microcosm
and the total shoot dry weight (D.W.) root dry D.W., and the percentage nitrogen (N)
of the dried shoot and root material recorded (Table 2).
Table 2. Total shoot D.W., root D.W. and N content (expressed as a %) of the shoots
and roots of the Lolium perenne plants grown in the presence of each of the organic
substrates added as discrete patches to the Lolium perennes root system.
Organic
Substrate Added
C:N Ratio
of substrate
Total Shoot
D.W. (g)
N in Shoots
(%)
Total Root
D.W. (g)
N in Roots
(%)
Shoot material 21:1 1.4864 1.32 0.7758 1.01
Root material 52:1 1.0614 1.09 0.6022 0.85
Urea 0.4:1 3.8441 1.48 1.3321 0.89
Amino acid
mixture 3.2:1 3.2354 1.25 1.2543 0.87
Algal cell
material 3.1:1 3.7447 1.09 1.1979 0.84
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From the data in Table 2
c) Calculate the total plant N content (as mg N) for the Lolium perenne plants grown
in the presence of each of the organic substrates. Please show all your calculations
and tabulate your answers. (7 marks)
d) Calculate the root weight ratio (RWR: root weight per unit total plant weight) forthe plants grown with each of the organic substrates. (4 marks)
Answer: (7 marks)
Organic Substrate
Added
Total Shoot N
content (mg N)
Total Root N content
(mg N)
Total Plant N content
(mg N)
Shoot material (1.4864 x [1.32/100])
x 1000 = 19.62
(0.7758 x [1.01/100])
x 1000 = 7.84
19.62 + 7.84 = 27.46
Root material (1.0614 x [1.09/100])
x 1000 = 11.57
(0.6022 x [0.85/100])
x 1000 = 5.12
11.57 + 5.12 = 16.69
Urea (3.8441 x [1.48/100])
x 1000 = 56.89
(1.3321 x [0.89/100])
x 1000 = 11.86
56.89 + 11.86 =
68.75
Amino acid mixture (3.2354 x [1.25/100])
x 1000 = 40.44
(1.2543 x [0.87/100])
x 1000 = 10.91
40.44 + 10.91 =
51.35
Algal cell material (3.7447 x [1.09/100])
x 1000 = 40.82
(1.1979 x [0.84/100])
x 1000 = 10.06
40.82 + 10.06 =
50.88
2ii) Root weight ratio (RWR): (4 marks)
RWR = root DW/ Total Plant DW (NB: No units)
Shoot material = RWR = 0.7758/2.2622 = 0.34
Root material = RWR = 0.6022/1.6636 = 0.36
Urea = RWR = 1.3321/5.1762 = 0.26Amino acid mixture = RWR = 1.2543/4.4897 = 0.28
Algal cell material = RWR = 1.1979/4.9426 = 0.24
As the organic substrates were labelled with the stable isotope of nitrogen (15
N) it was
possible to follow the fate of the N originally added in the organic substrates in the
different plant-soil-microbe pools (see Table 3).
Table 3.Percentage (%) of organic substrate N detected in plants, microbes and soil at
the end of the 55 days experimental period.Organic
Substrateadded
C:N
Ratio ofsubstrate
N from the organic
substrate in the L.perenne plants
(%)
Total
substrate Nlost from the
system
(%)
N from the organic
substrate in themicrobial biomass
(%)
Residual
substrate Nremaining
in soil
(%)
Shoot
material
21:1 11 12 15 62
Root
material
52:1 8 8 16 68
Urea 0.4:1 55 40 5 0
Amino acid
mixture
3.2:1 54 37 9 0
Algal cellmaterial
3.1:1 36 44 10 10
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e) The amount of N that the Lolium perenne plants captured from the organic patches
varied widely. Using the information in Table 3 describe in words what factors
controlled the amount of N that the plants captured from these organic substrates.
(2 marks)
Answer: (2 marks)
Main points are the C:N ratio of the substrate and the physical & chemical complexity
of the substrates.
The higher C:N ratio patches take longer to decompose as:
less N per unit C thus the micro-organisms require additional N for decomposition.
residual substrate values higher showing these patches still being decomposed.
micro-organisms control decomposition hence have turned over and released N forthe plants in the simple patches but N values lower in complex patches as bugs still
decomposing them.
Physical/chemical complexity:
C:N ratio of the amino acid mixture and algal cell wall material is similar yet plants
captured more N from the amino acid substrate as it is both physically and chemically
less complex than that of algal cell material thus would decompose and release N
more rapidly.
STUDY 3
In a field study close inspection of experimental plots suggested that the presence of
litter was associated with the presence of faecal pellets produced frommicroarthropods (small soil animals). To test this hypothesis, a grid of 10 x 50
contiguous area samples, each 25 cm2
was marked out. The presence or absence of
litter and faecal pellets was recorded for each sample. Of the 500 samples examined,
litter was present in 325, faecal pellets in 144 and both litter and faecal pellets were
present in 112.
f) (13 marks)
Show by conducting a statistical analysis how you would test the hypothesis that there
is an association between litter and faecal pellets. Please show all your calculations
including significance level, degrees of freedom and state the nature of the association
if there is one.
2
2 x 2 table (7 marks)
Litter Present Litter Absent
Faecal Pellets Present
Observed
Expected
112
93.60 (325 x 144)/500
32
50.40 (175 x 144)/500
144
Faecal Pellets Absent
Observed
Expected
213
231.40 (325 x 356)/500
143
124.60 (175 x 356)/500
356
325 175 500
Calculation of 2
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2 = (O E)2
/ E
2 = (112-93.60)2/93.60 + (32-50.40)
2/50.40 + (213-231.40)
2/231.40 + (143-
124.60)2/124.60
= 3.62 + 6.72 + 1.46 + 2.72 (4 marks)
= 14.52 with one degree of freedom giving P < 0.001
which indicates a highly significant association between litter presence and faecal
pellets, with the association being positive [as (ad) exceeds (bc)]. (2 marks).
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ECOLOGY 2
This question requires you to analyse data from field surveys of butterflies in
grassland and woodland sites. You examine whether different types of butterfly
species differ in their abundance between habitats. You are asked to critically
evaluate the methods used by the student in their surveys, and to suggest newfield work.
An ecology PhD student is investigating changes in the abundance of butterflies in
Britain. In the first year of the PhD studies, the student collates existing data on
butterfly abundance from four sites in Yorkshire. These data were collected by other
researchers using standardised methods.
The survey method involves an observer walking along a transect at a steady pace
counting all butterflies seen. Surveys are carried out only in sunny weather when the
temperature is above 17oC. Transects vary in length but are always 5m wide (i.e.
butterflies are recorded 2.5m either side of the observer). Transects were in two
different habitats; sites 1 and 2 were in woodland, and sites 3 and 4 in grassland.
Table 1 shows data collected from the two woodland sites in 2005.
Table 1. The total numbers of butterflies seen along two transects in woodland
sites in Yorkshire in 2005. The two transects were surveyed by a different person
but on the same day (20th
July). The length of each transect is shown. The eight
species that develop through several generations per year are marked with an
asterisk (all other species develop through a single generation per year).
Species Site 15.6 km Site 24.3 km
Species A* 7 8
Species B* 3 16
Species C 12 6
Species D* 49 17
Species E* 47 62
Species F* 39 53
Species G* 2 5
Species H* 1 0
Species I* 3 6
Species J 1 0Species K 14 9
Species L 2 2
Species M 46 19
Species N 6 4
1) The student is interested in investigating differences in butterfly abundance among
sites.
A) They decide to calculate the density per hectare (ha) of each species of butterfly at
each site. Table 2 below gives these density values for each species at the twograssland sites, as well as the mean density for this habitat. In a similar way, calculate
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A good answer would show that the data in Table 2 and the table of woodland means
had been correctly interpreted. For example, there are apparently higher abundances
of species in the grassland habitat compared with woodland. Species I has the highest
density at any site (site 4). Species H has the highest density in grassland but the
lowest density in woodland in other words, abundances do not seem to be positivelyassociated between habitats. Some species show much higher abundances in one
habitat than the other i.e. there is a suggestion of habitat specialisation
2) The student then decides to investigate differences in the ecologies of the species.
The student was aware that the species differ in the number of generations they
develop through each year. In Table 1 above, species that develop through several
generations per year (multivoltine species) are marked with an asterisk, all other
species develop through a single generation per year (univoltine). The student
calculated the mean density of each species across all sites and then carried out a
statistical test to determine whether multivoltine species have higher mean densitiesthan univoltine species. The SPSS output from this test is shown below.
Group Statistics
6 3.5181 3.33450 1.36130
8 13.8151 4.68888 1.65777
GENSunivoltine
multivoltine
TOTN Mean Std. Deviation
Std. Error
Mean
Independent Samples Test
.022 .886 -4.563 12 .001 -10.2970 2.25652 -15.21353 -5.38048
-4.800 11.990 .000 -10.2970 2.14508 -14.97114 -5.62287
Equal variances
assumed
Equal variances
not assumed
TOT
F Sig.
Levene's Test for
Equality of Variances
t df Sig. (2-tailed)Mean
DifferenceStd. ErrorD if fer enc e L owe r U pp er
95% ConfidenceInterval of the
Difference
t-test for Equality of Means
What statistical test did the student carry out? Explain what did they found. (4 marks)
Output shows that the student did a t-test comparing 6 univoltine species and 8
multivoltine species. There was a significant difference between these two groups (t =-4.56, 12 df, P = 0.001). Multivoltine species have significantly higher abundances.
Do you think the student did the appropriate statistical test? Explain your answer. (3
marks)
This is a parametric test and so the student should have tested for normality of data
beforehand. Levenes test shows no significant difference in variances and so gives
some support for a t-test. If the data were not normal, an alternative test for non-
parametric data would be a Mann-Whitney test.
3) The student was concerned that all these analyses were based on data from a single
transect at each site in July.
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A) How might this affect the findings? (5 marks)
Grassland species apparently have higher density but this might be because the survey
date was closer to their average peak flight period. Grassland species may comprise a
greater proportion of summer-flying species. There may be more nectar flowers in
grassland during July. Multivoltine species are apparently more likely to be grasslandspecialists this makes it difficult to separate the effects of these two aspects of the
species ecology on density. Multivoltine species may achieve even higher densities
than quoted if they develop through more generations later in the summer. This might
show if the transects were to be repeated later in the year. Single transect location
lacks spatial & temporal replication
B) Describe a series of observations or a field experiment that you could carry out to
test these ideas (5 marks)
Carry out transects in both habitats throughout the spring to autumn flight periods.
Determine timing of peak flight period for each species at the sites. Measure changesin flower abundance in the two sites over time. To get full marks you need to give
details on sample sizes, types of data to be collected and how they would be analysed
to test specific hypotheses.
4) The PhD student was particularly interested in changes in the abundance of one
butterfly species and collected data from surveys at Site 1 over the past 15 years. The
number of butterflies recorded on the transect each year is shown in Table 3. The
number of larval host-plants at the site was recorded as the mean percentage cover of
plants in 20 1m x 1m quadrats. Data for temperature and rainfall were obtained from
a meteorological station nearby.
Table 3. The Table shows changes at Site 1 in the total number of butterflies
recorded on the transect and the abundance of their larval host-plants over time.
year Total number of
butterflies recorded
%cover of larval host-
plants
1990 0 60.1
1991 0 9.2
1992 1 44.1
1993 0 30.1
1994 0 28.21995 1 62.1
1996 3 5.2
1997 5 2.8
1998 7 25.4
1999 11 10.1
2000 18 50.3
2001 27 43.7
2002 42 9.2
2003 45 17.2
2004 55 22.1
2005 49 25.7
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The student decides to analyse their data by using stepwise multiple regression
analysis to investigate whether host-plant abundance was affected by summer rainfall
or temperature in either the current year of study, or in the previous year. The SPSSoutput from this analysis is shown below. Data for host-plant abundance were arcsine-
transformed prior to analysis.
Model Summary
.935a .875 .865 .07839
Model1
R R Square
Adjusted
R Square
Std. Error of
the Estimate
Predictors: (Constant), RAINFALL PREVIOUSa.
Coefficientsa
-8.05E-02 .044 -1.837 .089
3.712E-02 .004 .935 9.543 .000
(Constant)
RAINFALL
previous
Model1
B Std. Error
Unstandardized
Coefficients
Beta
Standardized
Coefficients
t Sig.
Dependent Variable: ARCSIN PLANT COVERa.
Excluded Variablesb
-.102a
-1.039 .319 -.287 .985
-.068a
-.671 .515 -.190 .976
.068a
.673 .514 .191 .985
RAINFALL
current
TEMP
current
TEMP
previous
Model1
Beta In t Sig.
Partial
Correlation Tolerance
Collinearity
Statistics
Predictors in the Model: (Constant), RAINFALL PREVIOUS YEARa.
Dependent Variable: ARCSIN PLANT COVERb.
A) Are any of the factors significant in the analysis? If so, list the significant
factors(s). (1 mark)
Yes, rainfall previous year
B) What is the value of the slope of the significant relationship? (1 mark)
Slope = 0.0371
C) Why were data for plant cover arcsine transformed prior to analysis? (1 mark)
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Plant cover was scored as a percentage (i.e. a proportion). The variance of a
proportion is a function of the mean. This is undesirable in statistical analyses.
Arcsine transformation removes (or, at least, reduces) this dependence.
D) Explain the R square (R2) value. (1 mark)
The model explains 88% of the variation in the data set.
E) Discuss the findings of this statistical analysis, and whether this was the most
appropriate way to analyse the data. (3 marks)
Plant cover is affected only by rainfall in previous year, not rainfall in current year.
No relationship with any temperature data. The lag response of a year indicates the
host plant may be an annual. The relationship with rainfall indicates the site is dry
such that plants are water-limited.
Data from consecutive years are unlikely to be independent (as assumption of the
regression analysis), and so stepwise regression may not be the best approach. A time-series analysis might be better.
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GENETICS 1
This question is about genetic mapping in bacteria (which has been incredibly
important in the development of molecular biology) using conjugation, in which
the circular chromosome of the donor cell is broken at a fixed point and
transferred as a linear piece of DNA to a recipient.
A. Interrupted mating is a technique that is used for long distance mapping of genes
on the E. coli chromosome. A donor strain transfers its chromosome linearly from a
fixed point on the circular genome into a recipient strain that usually contains multiple
mutations. Samples are removed from the mating culture at various time points after
mixing the two strains and transfer of the DNA is halted. This is called interruption
of mating. The bacteria in the interrupted sample are plated out on various selective
media that will kill the donor strain and that lack a single requirement of the recipient
strain, so that one can determine if, at the time of interruption, the recipients have
inherited a particular gene. For example if the donor was arg+leu
+pro
+thr
+
streptomycin sensitive and the recipient arg- leu- pro-thr- streptomycin resistant, one
would plate out the interrupted mating mix on four selective media. The selective
plate for arg+
recombinants would contain leucine, proline, threonine and
streptomycin but no arginine. Note that for the gene to be inherited, not only does it
have to be transferred to the recipient but it has to be recombined into the recipients
chromosome.
In an undergraduate class practical on interrupted mating, the donor was arg+leu
+
pro+thr
+and the recipient arg
-leu
-pro
-thr
-. When the organiser ran through the
procedure before the practical, he obtained some bizarre results, that he interpreted
correctly to mean that none of the selective plates contained leucine. However therewas no time to make the plates again, so he decided to go ahead with the experiment
anyway. The times of entry he had expected if the plates had been correct, were pro+
at 10 minutes after mixing, leu+
at 18 minutes, thr+
at 20 minutes and arg+
at 30
minutes.
1) What times of entry for each gene would the class observe? (3 marks)
pro+ 18 leu+18 thr+20 arg+30
If two genes enter more than 5 minutes apart they behave as though they are unlinked
i.e. the recombination frequency (RF) between them is 50%. The RF between leu+
and thr+
is 30%.
2) Write down the percentage of a) leu+
b) thr+, c) arg
+colonies observed by
the class at late times (e.g 60 minutes), taking as 100%, the number they
should have observed if the plates had been correct. (2 marks)
a)Leu+ 100 b) thr+70 c) arg+ 50
B. A bacterial geneticist investigating the rotary motor that drives the flagellae of
E.coli, isolates a spontaneous non motile mutant in the donor strain used in part A,
and attempts to map it using conjugation. This time the recipient strain is pro-
(10min) leu- (18min) thr- (20min) arg- (30min) ilv- (34)min aro- (48min). For this
mapping she cannot use interrupted mating to give time of entry of non-motilitybecause there is no easy way to select for it. So instead she uses recombination
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frequencies. She allows the donor and recipient strains to conjugate for two hours
and then plates 0.1ml of serial tenfold dilutions of the mixture out on the six different
types of selective plate (which this time are correctly made up) and counts the
colonies that appear on the plates. She then calculates the number/ml of each type of
recombinant in the original mating mix. The numbers she obtains are:
pro+
6.0*107/ml
leu+
3.0*107/ml
thr+ 2.9*107/ml
arg+
1.2*107/ml
ilv+
9.4*106/ml
aro+
3.2*106/ml
3) Given that when counting bacterial colonies on plates there should be at least
50 and no more than 500, what dilutions of the mating mix should she have
spread for each of the different recombinant types. (3 marks)
In order from the top 105, 104, 104, 104, 104, 103
4) If the logarithm of the numbers of recombinants obtained are plotted on the y
axis versus time of entry on the x axis a straight line is obtained. Comment
on the relationship between time of entry and numbers of recombinants
obtained. (4 marks)
On log graph paper it is a straight line indicating exponential decay or a constant
probability per unit time (or unit distance of transfer) that transfer of the genes will
cease. Called spontaneous interruption of mating
She then mapped the mutation that confers non-motility by laboriously picking 100 of
each type of recombinant, growing them up and checking for mobility with a
microscope. She obtains the following data.
Recombinants Number non-motile (out of 100)
pro+
9
leu+
16
thr+
19
arg+ 60
ilv+
70
aro
+
50
From these data she concludes that the mutation conferring non-motility is exactly
halfway between ilv+
and arg+.
5) Explain why there are fewer arg+
recombinants that are non-motile than ilv+
recombinants that are non-motile if the mutation is exactly equidistant from
each. (4 marks)
Not all the arg+
recombinants will have received the non-motility mutation while all
the ilv+
ones will have so that with the same recombination frequency there will be
less arg+
mot-
than ilv+
mot-.
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Now that she has mapped the defective gene she looks for candidate genes of
unknown function in that place on the published genome of E.coli. She finds a
cluster of five Open Reading Frames that have been designated as probable genes, all
oriented in the same direction.
6) What is an Open Reading Frame and what property/properties would indicatethat one is likely to be a gene? (3 marks)
An ORF is a sequence without stop codons. In bacteria the length is the most
important property. Should be 100 or more. Other criteria would be a start codon
and Shine Dalgarno sequence
Close examination of the sequence containing the five ORFs suggests that it contains
only a single promoter at the appropriate end and therefore that the five genes make
up a single transcription unit or operon. The geneticist scans the sequence for
restriction sites, chooses an enzyme that will not cleave within the sequence but does
so at appropriate distances beyond the ends, and uses it to clone the fragment into a
plasmid. Having assured herself by partial sequencing that she has the appropriatefragment she inserts it into the original non-motile strain and finds that the bacteria
are now motile.
7) What does she conclude? (2 marks)
That the plasmid contains the wild type version of the defective gene(s) and the wild-
type genes are dominant to the defective ones
To find out which gene(s) the mutation is located in she separately clones each gene
present in the fragment into a suitable expression vector (i.e. one that allows high
expression of a cloned gene on addition of a suitable inducer) and puts each of the
new constructs into the original non-motile strain. After addition of inducer she finds
that none of the individual genes confers motility on the strain.
8) What can she conclude about a) the function of the genes in the operon and b)
about the effect of the mutation? (6 marks)
a) That the original large 5 gene fragment contains more than one gene required for
motility
b) that the mutation inactivates more than one gene.
9) The mutant and wild-type operons are sequenced and the mutation turns out to be
a single base insertion in the coding part of the first gene. Explain how such a
mutation might cause the phenotype of the mutant. (6 marks)
The mutation is a polar mutation. It inactivates genes in the same operon downstream
of the mutation. The frame shift causes the ribosomes to terminate at a previously
out-of-frame termination codon. This would leave the RNA polymerase to
synthesise a section of RNA that is not covered by ribosomes. This in turn could lead
to termination of transcription (the usual reason for polarity) or there might be a
translational coupling effect where ribosomes must translate the first gene copied intothe mRNA to allow the second gene copy to be translated and so on. It is thought
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that the ribosome translating the early gene copy unravels secondary structure in the
message thus exposing the next translational start signal for binding by the same or a
new ribosome. Another explanation might be that the mRNA is destabilised by
having no ribosomes protecting it over a section of the first gene (between the out of
frame stop and the beginning of the second gene) and this causes very rapid
degradation of the mRNA for the last four genes. This is less likely than the firsttwo explanations but is a plausible idea.
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GENETICS 2
This problem asks if the difference between alleles of a mutant locus is
significant and if so, what might be the molecular explanation. Mutational
changes in the DNA sequences another mutant locus are shown and molecularexplanations for their effect are required.
Question 1A
In Drosophila, the recessive mutation hirsute (hir) causes scutellar bristle length to
increase by approximately 200%
There are 2 bristles per individual fly measured.
n = number of bristles measured
n avg. bristle length (m) standard deviation
wild type 20 89.29 22
hir1/hir
220 178.35 41
hir1/+ 20 118.5 31.5
hir2/+ 20 123.25 24.6
hir1/+ and hir
2/+ have been included as controls. However the hir
1/+ and hir
2/+
heterozygotes appear to have longer bristles than wild type (wt).
Calculate the standard errors (1 mark)
s.e. wt 4.92
hir/hir 9.17
hir1 7.04
hir2 5.50
Estimate if the differences are likely to be significant between wild type and the hir
heterozygotes (1 marks) and explain your reasoning (1 mark)
With df= 38 for each test, the t value must be greater than t=2.04 (P=0.05) for there to
be a 95% chance of the means being significantly different. A 95% chance of
significant difference between wt and hir1/+ (t=3.39) and wt and hir2/+ (t=4.6)
What are two possible genetic explanations for this result? (4 marks)
Dominance or haploinsufficiency
After molecular analysis, hir1and hir
2alleles are found to be null alleles. Which
genetic explanation does this information favour? Explain your reasoning. (4 marks)
Haploinsuffuciency. Nulls produce no transcript or protein so cannot produce
dominance, so insufficient gene product produces a phenotype
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Question 2
The sideparting (spar) gene has been cloned and a cDNA isolated (below). The
cDNA encodes a short protein, the start ATG codon is in lower case and the stop
TGA codon is in lower case.
TATAAGCATCCGATCCAACCCGAACCGATCatgGCAACCACTCCACGCAGCGGCGGT
AA
GTTCGAGATCTGGGACACGGCTGGCCAGGAGCGGTACCACAGCTTAGCTCCCATGTA
TT
ATCGAGGAtgaGAAAATGAAGGAAAACGAAAACCACAAAAAAAAAAAGAAAACCAA
A fragment of the genomic region of the spar gene in the spar1
mutant was
sequenced. The underlined G was mutated to an A. By comparing the cDNA with thegenomic sequence suggest what the molecular genetic consequence of such a
mutation would be? (10 marks)
In frame stop codon generating a truncated transcript. Either producing a truncated
protein or transcript is degraded by nonsense mediated decay process.
TATAAGCATCCGATCCAACCCGAACCGATCATGGCAACCACTCCACGCAGCGGCGGTAAGTT
CGA
GATCTGGGACACGGCTGGCCAGGAGCGGTAAGTATCGCTGGATAGATCACCCAACTGAAAGC
TTC
ATCTGACATACTTATATTCGCTTTTGTAGGTACCACAGCTTAGCTCCCATGTATTATCGAGG
ATG
AGAAAATGAAGGAAAACGAAAACCACAAAAAAAAAAAGAAAACCAA
In the spar2
mutant stock the genomic region of the spargene was sequenced. Three
nucleotides were found to be missing (underlined). By aligning the cDNA sequence
with the genomic sequence, suggest what type of molecular genetic defect the three
missing nucleotides would generate (12 marks).
TATAAGCATCCGATCCAACCCGAACCGATCATGGCAACCACTCCACGCAGCGGCGGTAAGTT
CGA
GATCTGGGACACGGCTGGCCAGGAGCGGTAAGTATCGCTGGATAGATCACCCAACTGAAAGC
TTC
ATCTGACATACTTATATTCGCTTTTGTAGGTACCACAGCTTAGCTCCCATGTATTATCGAGG
ATG
AGAAAATGAAGGAAAACGAAAACCACAAAAAAAAAAAGAAAACCAA
The three nucleotide deletion results in removal of bases that constitute a spliceacceptor sequence, generating a nonsense sequence that would produce either a
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scrambled protein, an early truncation, or reduced transcript due to nonsense mediated
decay.
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CELL AND ORGANISMAL BIOLOGY 1
Investigating the characteristics of the eWnt secreted signaling molecule. Analysis and
manipulation of DNA plasmid constructs coding for eWnt protein. Analyzing the
properties the eWnt protein and the eWnt gene promoter in vitro and in vivo.
Wnt proteins are secreted glycoprotein ligands. Binding of Wnt ligands to their cell surface
receptors activates an intracellular signaling pathway which plays a critical role in regulating
gene expression in the developing embryo. The downstream transcriptional effectors of the
Wnt pathway are TCF/Lef transcription factors which bind sites matching the consensus
ATCAAAG in the promoter regions of Wnt target genes. This problem paper investigates
embryonic Wnt (eWnt); a novel member of the Wnt family of ligands that has been
identified in the frog Xenopus tropicalis. The single large open reading frame (ORF) codesfor a primary translation product of 40 Kd. A restriction enzyme map of the eWnt cDNA is
shown in Figure 1.
Experiments designed to look at the properties of the eWnt protein make use of thepTran vector. Figure 2 details the multiple cloning site region of the 3400 bp pTran vector.
The restriction sites indicated are for enzymes which cut only once within the pTran vector.
Sub-cloning the eWnt cDNA in an appropriate orientation into the BamHI site of pTran
allows transcription of a synthetic mRNA coding for the full length eWnt protein using the
T7 phage RNA polymerase promoter.
Figure 1 Map of the eWnt cDNA
Figure 2 Multiple clones site region of pTran plasmid
Please show all relevant working and calculations.
Question 1: (2 marks)
How many amino acids are in the conceptual eWnt protein?
367 amino acids
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Question 2: (2 marks)
The 1225 bp fragment containing the eWnt ORF is sub-cloned into the BamHI site of the
3400 bp pTran vector. The ligation reaction to sub-clone the eWnt ORF fragment into pTran
requires a 3:1 molar ratio of insert DNA to vector DNA. 100 ng of pTran vector DNA is
used in the ligation reaction. Assuming that the average base content in both vector and
insert is the same, calculate the mass of insert DNA (in ng) which will be added to thereaction in order to give the 3:1 molar ratio of insert to vector DNA?
108 ng
Question 3: (1 marks)
A clone containing the ORF fragment in the correct orientation is identified. In order to
produce a suitable transcription template the circular pTran-eWnt plasmid DNA must be
linearised to enable the production of a run off RNA transcript. Which restriction enzyme
must be used to produce the linear template required for the sythesis of a run off transcript
containing the whole of the eWnt ORF and the poly adenylation signal sequence?
NotI
Question 4: ( 3 marks)
The eWnt mRNA has been transcribed and purified. The purified product is contained in a
final volume of 20 l. A 1/1000 dilution of the product has an OD 260 of 0.014. 40 g/ml
RNA has an OD of 1. What is the total mass (in g) of the mRNA product?
11.2 g
Question 5: (3 marks)
A cell lysate based in vitro translation system is used to investigate the processing of the
eWnt protein. The addition of synthetic eWnt mRNA to the cell lysate results in de novo
translation of eWnt protein. This assay requires the use of 1 g of mRNA per reaction.
What volume of your transcription product will you need to use for each translation reaction?
1.79 l
Question 6: (5 marks)
The sizes of translation products are analyzed by SDS-PAGE. As predicted the size of the
primary translation product is about 40 Kd. Endoplasmic reticulum vesicles purified from
canine pancreas when added to the in vitro translation reactions allow post-translationalmodification of secreted proteins. The addition of ER vesicles to the reactions leads to a
shift in size of the product to 48 Kd. Why does this shift in size occur?
Wnt proteins are secreted glycoproteins which will normally enter the secretory
pathway by translocation into the ER. Once in the ER the protein can by
glycosylated. Translation in the absence of ER vesicles gives the expected size of the
unprocessed primary peptide. Addition of the ER vesicles allows translocation of the
primary into the vesicles lumen where they can be processed. Glycosylation of the
Wnt protein causes an increase in molecular weight to 48 KDa.
Question 7: (5 marks)
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When the translation product produced in the presence of ER vesicles is subsequently treated
with the enzyme N-glycosidase the size of the product shifts to 38 Kd.
Why is the product now smaller than the predicted primary product size of 40 Kd?
The N-glycosidase enzyme removes the glycosylation from the product synthesized in the
presence of the ER vesicles causing a reduction in the size. The resulting product is smaller
than 40 KDa because during translocation into the ER the signal peptide is cleaved from theamino terminus of the protein.
Question 8: (6 marks)
A region upstream of the transcriptional start site of the eWnt gene has been isolated and
cloned upstream of a minimal eukaryotic promoter designed to drive expression of the
luciferase enzyme based reporter gene. This construct is termed eWnt luc.
Figure 3A shows the profile of the relative levels of expression from the endogenous
Xenopus eWnt gene at a number of time points post fertilization. Figure 3B shows the
profile of the relative levels of reporter activity in Xenopus embryos carrying the eWnt-luc
transgene.
What do these data tell us about the eWnt-luc transgene and the upstream sequencethat it contains?
Figure 3A Figure 3B
The temporal expression profile for the endogenous gene is closely matched by that of the
reporter transgene. This indicates that the reporter is able to recapitulate the expression of
the endogenous gene. Therefore the identified upstream region of eWnt within the eWnt-
luciferase transgene contains all the regulatory information necessary to drive the normal
expression pattern of this gene.
Question 9: (6 marks)
In subsequent experiments eWnt protein is overexpressed from injected mRNA in embryos
carrying the eWnt-luc transgene and in embryos carrying modified eWnt-luc transgenes inwhich either bases 1 to 30 or bases 31 to 54 have been deleted from the upstream region
(eWnt-delta30-luc and eWnt-delta54-luc respectively). The sequence of the complete
upstream region of eWnt-luc is shown in Figure 4A. Figure 4B shows reporter activity in
embryos carrying these transgenes in the absence or presence of overexpressed eWnt protein.
Present an hypothesis to explain the observed effects on reporter activity.
Examination of the sequence 31 to 54 reveals the presence of 2 consensus TCF
binding sites for TCF/Lef transcription factors. These are likely to be involved in
mediating the reporter response to Wnt signaling. Removal of these sites renders the
reporter unresponsive to Wnt signaling.
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Figure 4A Upstream region of eWnt
Figure 4B
Reporter expression
0
100
200
300
400
500
600
eW nt -luc eW nt-
luc+eWnt
protein
eWnt-delta
30-luc
eWnt-delta
30-luc+eWnt
protein
eWnt-delta
54-luc
eWnt-delta
54-luc+eWnt
protein
Relativeexpression
Reporter expression
0
100
200
300
400
500
600
eW nt -luc eW nt-
luc+eWnt
protein
eWnt-delta
30-luc
eWnt-delta
30-luc+eWnt
protein
eWnt-delta
54-luc
eWnt-delta
54-luc+eWnt
protein
Relativeexpression
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CELL AND ORGANISMAL BIOLOGY 2
This question concerns the use of oxygen by diving mammals. The first part of
the question is to investigate the relation between metabolic rate and size, and
how this may be influenced by their diving habits. The second part of the
question concerns the use of oxygen by human divers.
Table 1
Type of animalLive mass of
animal (kg)
Rate of use of
oxygen (ml / min.)
Oxygen stored by
the animal (ml / kg)
Heart beats / minute
At surface | Diving
Common Seal 100 170 46 92 14
Grey Seal 500 560 52 51 8
Walrus 1 100 1 000 46 40 unkn.
Lesser Rorqual 5 000 3 150 52 27 unkn.
Humpback Whale 22 500 9 700 50 18 3
Greenland Whale 66 000 21 800 45 14 unkn.
Human 70 250 8 80 50
Horse 650 1800 unkn 35 -
Elephant 3800 4460 unkn 28 -
unkn. : Unknown
1. Mammals that dive under water at 10C use oxygen during submergence at
the rates shown in Table 1.
a. Draw graphs to show the relationship in Table 1 between the size of a
mammal, its oxygen consumption and heart rate. (9 marks)
a.
100
1000
10000
100000
10 100 1000 10000 100000
mass (kg)
rateofoxygenuse(m
l/min)
acquatic
terrestrial
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0
10
20
30
40
50
60
70
80
1 10 100 1000 10000 100000
mass (kg)
oxygencapacity(ml/kg)
acquatic
terrestrial
1
10
100
1.0 10.0 100.0 1000.0 10000.0 100000.0
mass (kg)
heartrate(beats/min)
acquatic surface
terrestrial surface
acquatic dive
terrestrial dive
Summarise in words the differences in oxygen consumption between acquatic and
terrestrial animals, and the effects of diving on oxygen consumption. (9 marks)
Rate of oxygen use is a log log function of mass (very good straight line)
Terrestrial mammals above their acquatic friends (though the whales etc live in colder
water, energetically easier to move through water than on land)
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Oxygen capacity is independent of size, and is much more than humans. Need this to
store oxygen to use while diving
Heart rate declines in log-log fashion with size, and it appears that the terrestrial
animals have a lower heart rate at the surface
During diving, heart rate reduces (bradycardia), very substantially greater drop for
acquatic than terrestrial
2. Suggest why the rates of heart beat, where known, are so low when diving
compared with values at the surface and any other aspect of physiological
adaptation associated with this. (5 marks)
While diving, no air outside and probably not much oxygen will be in lungs
(compressed by pressure) so dont need heart to pump blood around the
muscles to transfer oxygen from air to muscle. Most oxygen probably storedin muscle myoglobin anyway,
3. Human divers commonly breathe compressed ordinary air (21% O2, 78% N2and 1% Argon) when diving in shallow water. The hydrostatic pressure in
water increases steadily by 1 atmosphere for each 10 metres below the surface
a. How does the partial pressure of oxygen increase with depth? (2
marks)
linearly, at 0.2 atm for each 10 m
b. What would you expect the partial pressure of oxygen to be in the
human body at 70 metres depth (show calculations)? (6 marks)
partial pressure at 70 m = 0.2 * (70 / 10 * 1 (for water) + 1 (for
atmosphere) atm ) = 1.6 atm
c. Oxygen is quickly toxic at a partial pressure of about 2 atm. At what
depth would this be reached? (2 mark)
if d is depth;
2 = 0.2 * (d / 10 * 1 (for water) + 1 (for atmosphere) atm )
2 = 0.2 * d /10 + 0.2
2 0.2 = 0.02 * d
1.8 / 0.02 = d = 90 m
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MOLECULAR BIOLOGY AND BIOCHEMISTRY 1
Loss of function mutations were generated on plasmids and investigated using
yeast cultures. Measurements were made so that the effect of the mutations
could be determined and hypotheses devised to explain the observations.
Two point mutations, Mut1 and Mut2 were generated in gene X in separate copies of
shuttle vector plasmids based on bioinformatics data that suggested that these
mutations would result in loss of function mutations. An experiment was designed to
measure the rate at which plasmids containing the Mut1 and Mut2 mutations were lost
from the yeast Saccharomyces cerevisiae. Yeast cells were transformed with the
constructs outlined in Figure 1. The plasmids possessed a centromere sequence
(CEN), which allows them to exist at low copy number but be accurately segregated
at mitosis and an Autonomously Replicating Sequence (ARS) that allows replication
of the plasmid in yeast. The URA gene encodes a protein required for uracil
production that is missing in the host yeast strain. S. cerevisiae cells containing the
plasmids were selected for on complete medium plates that did not contain uracil.
Colonies were inoculated into liquid cultures without uracil and grown into the log
phase. Aliquots of these cultures were inoculated onto plates with and without uracil,
and the liquid culture was used to inoculate another liquid culture that DID contain
uracil. The cells were grown for a further four generations and then two further
aliquots were plated onto media with and without uracil. The plates were incubated
until colonies were visible, and the number of colonies on each plate was counted
(Table 1).
Table 1
Plasmid
No. of colonies before growth in
non-selective media
No. of colonies after 4
generations without selection
media media-ura media media-ura
WT 1189 1322 2055 2263
Mut1 467 504 1521 489
Mut2 904 655 1505 441
1. Calculate the stability of each plasmid under non-selective conditions using theformula (6 marks):
% cells that lose plasmid per generation X= (1 er) 100
Where: rln(
A
B)
N
and A=% cells containing plasmid after Ngenerations without selection, B=% cells
containing plasmid before non-selective growth
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WT= 0.3% cells that lose plasmid/ generation
Mut1= 26% cells that lose plasmid/ generation
Mut2=22.9% cells that lose plasmid/ generation
2. Identify which of the three plasmids is most stable and which is most unstable (5
marks).Most stable is WT with 0. 3% of cells losing plasmid/generation and most unstable is
Mut1 with 26% loss.
The plasmids used to generate the plasmid stability data were digested with HindIII,
SmaI and EcoRI. The resulting DNA fragments were separated on an agarose gel,
stained with ethidium bromide and visualised under UV light. The results are shown
in Figure 2.
3. Construct a plasmid map for the Mut1 plasmid (10 marks).
Answer:
Hind III 1625
Amp
Xba I 5050
0
Bam HI 900
Eco RI 580
Eco RI 2690
Sma I 2220
URA 3
Gene X
pCEN-Gene X-URA5450 bp
CENARS
ori
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4. Suggest a hypothesis for why the Mut1 and Mut2 plasmids are unstable (6 marks).
Mut1: the effect likely to be due to loss of part of CEN and not Gene X. Or Mut1 is a
dominant negative of a gene required for plasmid maintenance.
Mut2: the effect is due to a mutation in the Gene X, (dominant negative effect).
5. In the case of Mut1 how could you test your hypothesis (6 marks)?
Mut1: Put the WT Gene X in the CEN deletion plasmid and Mut1 Gene X in the full-
length plasmid to separate the effects of the Mut1 mutation from the effect of only
having a partial CEN.
Figure 1. Map of plasmids used to investigate the function of gene X in yeast. Three
plasmids were generated and tested. The plasmids possessed either the WT Gene X
or one of two mutations believed to result in loss of function.
0 Eco RI 580
Amp
Xba I 6120
Eco RI 5090
ori
Bam HI 4351
Sma I 4060
Bam HI 900
Eco RI 2690
Sma I 2220
Hind III 1625
URA 3
Gene X
pCEN-Gene X-URA
6500 bp
CEN
ARS
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Figure 2. Restriction enzyme digestion of plasmids recovered from yeast.
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MOLECULAR BIOLOGY AND BIOCHEMISTRY 2
Overexpression of protein at different temperatures: this problem concerns the
synthesis of a specific protein in a bacterial expression system. It deals with
analysis of bacterial growth, and characterisation of the product by gel
chromatography.
Introduction
During overexpression of recombinant proteins in Escherichia coli, misfolded
proteins may aggregate and form insoluble inclusion bodies. One factor that can
affect the tendency of a bacterially overexpressed protein to aggregate is the
temperature of cell growth. Rapid growth at 37 C leads to high rate of protein
expression, and generally a greater tendency of the protein to aggregate and form
inclusion bodies than when cells are grown at lower temperature, where the rate of
protein synthesis is slower. The aim of this study was to assess the effect oftemperature on the rate of protein expression and it solubility
A plasmid coding for the protein under study was transformed into a suitable
Escherichia coli strain and grown at 37 C in LB medium containing 100 g/ml
ampicillin. Cell density was monitored, and when the value of A600 was 0.2 protein
expression was induced by the addition of IPTG. At this point the culture was divided
into two equal portions, one of which was cooled to 25 C and the other was left to
grow at 37 C. Measurement of cell density was continued, and the results are shown
in Table 1. Samples were taken from the 37 C culture at 0, 1, 2 and 3 hours after
induction, and from the 25 C culture at 0, 1 , 3 and 5 hours after induction.
Time A600(mins) 37 C 25 C
0 0.04
30 0.038
60 0.042
90 0.048
120 0.05
150 0.1
180 INDUCTION 0.2 0.2
210 0.4225 0.28
240 1 hr post-induction 0.8
270 1hr post-induction 1.6 0.4
300 2 hr post-induction 1.7
315 0.56
360 3 hr post-induction 1.8
390 1.03
480 5 hr post-induction 1.8
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Table 1. Cell density at 600 nm (A600) for growth at 37 C and 25 C. Note that cells
were grown at 37 C for 3 hours before the cultures were divided for growth at the
two temperatures.
Question 1 (12 marks 4 for each part) Using an appropriate form of graphicalplot evaluate:
(i) the lag time before the onset of logarithmic growth at 37 C
Lag time = 115 mins
(ii) the doubling time for the cell cultures at 37 C and 25 C
Doubling time at 37 C = 30 mins
(iii) stating any assumptions, estimate the value of A600 for the 25 C sample taken at
3 hours after induction of protein expression.
Doubling time at 25 C = 90 mins
Cells were harvested from 100 ml samples of the cell cultures grown for 3 hours at
37 C and 5 hours at 25 C (under the same conditions as Table 1, Part 2), lysed by
sonication, and centrifuged to yield a supernatant fraction (S) and pellet fraction (P)
for each growth temperature. A sample of total cell extract (T) was also taken before
centrifugation.
Small samples (from equivalent cell densities) were taken for protein analysis by
denaturing polyacrylamide gel electrophoresis (SDS-PAGE). All three samples (T, P
and S) from each growth temperature were denatured by boiling in SDS. The resulting
gels were stained for total protein with Coomassie blue, and the results are shown
diagrammatically in Figure 2.
T S P T S P
A (37 C) B (25 C)
(A) (B)
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Figure 1. SDS-PAGE gels of the total cell protein(T), and the soluble (S) and pellet
(P) fractions of expressed protein from (A) the 37 C cell growth and (B) the 25 C
cell growth cultures. The protein MW ladder is shown on the left of each gel.
Question 2 What conclusions do you draw from the SDS-PAGE results about theform of the expressed protein at the two temperatures? (5 marks)
There is a lot more soluble protein a produced at 25 C than at 37 C, as evidenced bythe band intensities
Samples of soluble protein fractions from 100 ml of the 37 C and 25 C cell cultures
were dialysed for purification on a Ni-nitriloacetate (Ni-NTA) column, which binds
His-tagged proteins. The his-tagged protein proteins were eluted by imidazole
gradients (in separate experiments) as pure protein, as demonstrated by SDS-PAGE.
The total amount of pure protein from the 37 C cells was 3.0 mg, and from the 25 C
cells it was 25.2 mg.
Question 3 From the total protein yield from the Ni-NTA column, calculate the
concentration of soluble protein per ml of cell culture in the 37 C and 25 C cultures.
(10 marks)
At 37 C 3.0 mg soluble protein were produced from 100 ml culture. Therefore
concentration of soluble protein per ml = 30 g/ ml
At 25 C 25 mg of soluble protein from 100 ml, so the concentration is
250 g/ml
Question 4 Outline how would you quantitatively assess the proportion of protein that
was produced in soluble form at the two temperatures? (6 marks)
By scanning the gels and estimating the proportion of soluble to total protein.