problem specimen assembled

7/31/2019 Problem Specimen Assembled

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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.

One the actual paper there will be four questions, one each for Ecology,Genetics, Cell and Organismal Biology and Molecular biology and

biochemistry.

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.

Short outline of each question:

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.

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

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)

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).


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Table 3.Percentage (%) of organic substrate N detected in plants, microbes and soil at

the end of the 55 days experimental period.Organic

Substrate

added

C:N

Ratio of

substrate

N from the organic

substrate in the L.

perenne plants

(%)

Total

substrate N

lost from the

system(%)

N from the organic

substrate in the

microbial biomass

(%)

Residual

substrate N

remaining

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 cell

material

3.1:1 36 44 10 10

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)

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 from

microarthropods (small soil animals). To test this hypothesis, a grid of 10 x 50

contiguous area samples, each 25 cm

2

was marked out. The presence or absence oflitter 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.


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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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density per hectare (ha) of each species for the two woodland sites and mean density

for the woodland habitat. You should tabulate your answers. (4 marks)

What might the student conclude about the relative distributions and abundances of

the butterfly species in the two habitats? (5 marks)

Table 2. The density per hectare (ha) of each species of butterfly at two grassland

sites, and the mean butterfly density in the grassland habitat.

Species Site 3

density

Site 4

density

Mean

density in

grassland

Species A 22.44 20.00 21.22

Species B 27.32 25.95 26.64

Species C .49 .00 .25

Species D 2.93 1.08 2.01

Species E 3.41 4.86 4.14

Species F 22.44 32.97 27.71

Species G 13.17 24.86 19.02

Species H 27.32 28.11 27.72

Species I 19.02 34.59 26.81

Species J .98 1.62 1.30

Species K .49 1.08 .79

Species L .00 .00 .00

Species M 1.46 3.24 2.35

Species N 10.24 17.30 13.77

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 densities

than 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


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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

assumedEqual variances

not assumed

TOT

F Sig.

Levene's Test for

Equality of Variances

t df Sig. (2-tailed)

Mean

Difference

Std. Error

D if fer enc e L owe r U pp er

95% Confidence

Interval of the

Difference

t-test for Equality of Means

What statistical test did the student carry out? Explain what did they found. (4 marks)

Do you think the student did the appropriate statistical test? Explain your answer. (3

marks)

3) The student was concerned that all these analyses were based on data from a single

transect at each site in July.

A) How might this affect the findings? (5 marks)

B) Describe a series of observations or a field experiment that you could carry out to

test these ideas (5 marks)

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 28.2

1995 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 SPSS

output 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

CollinearityStatistics

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 significantfactors(s). (1 mark)

B) What is the value of the slope of the significant relationship? (1 mark)

C) Why were data for plant cover arcsine transformed prior to analysis? (1 mark)

D) Explain the R square (R2) value. (1 mark)

E) Discuss the findings of this statistical analysis, and whether this was the most

appropriate way to analyse the data. (3 marks)


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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)

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)

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-motility

because there is no easy way to select for it. So instead she uses recombination

frequencies. She allows the donor and recipient strains to conjugate for two hoursand then plates 0.1ml of serial tenfold dilutions of the mixture out on the six different


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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

/mlthr

+2.9*10

7/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)

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. Commenton the relationship between time of entry and numbers of recombinants

obtained. (4 marks)

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 fromeach. (4 marks)

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 indicate

that one is likely to be a gene? (3 marks)

Close examination of the sequence containing the five ORFs suggests that it containsonly a single promoter at the appropriate end and therefore that the five genes make


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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 appropriate

fragment she inserts it into the original non-motile strain and finds that the bacteria

are now motile.

7) What does she conclude? (2 marks)

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)

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)


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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 1

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)

Estimate if the differences are likely to be significant between wild type and the hir

heterozygotes (1 marks) and explain your reasoning (1 mark)

What are two possible genetic explanations for this result? (4 marks)

After molecular analysis, hir1and hir

2alleles are found to be null alleles. Which

genetic explanation does this information favour? Explain your reasoning. (4 marks)

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


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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 the

genomic sequence suggest what the molecular genetic consequence of such a

mutation would be? (10 marks)

.

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


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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?



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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 the

reaction in order to give the 3:1 molar ratio of insert to vector DNA?


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?

Question 4: ( 3 marks)

The eWnt mRNA has been transcribed and purified. The purified product is contained in afinal 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?


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?


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-translational

modification 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?


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?


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 sequence

that it contains?


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Figure 3A Figure 3B


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 in

which 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 inembryos carrying these transgenes in the absence or presence of overexpressed eWnt protein.

Present an hypothesis to explain the observed effects on reporter activity.

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)

Summarise in words the differences in oxygen consumption between acquatic and

terrestrial animals, and the effects of diving on oxygen consumption. (9 marks)

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)

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)

b. What would you expect the partial pressure of oxygen to be in the

human body at 70 metres depth (show calculations)? (6 marks)


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c. Oxygen is quickly toxic at a partial pressure of about 2 atm. At what

depth would this be reached? (2 mark)


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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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2. Identify which of the three plasmids is most stable and which is most unstable (5

marks).

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).

4. Suggest a hypothesis for why the Mut1 and Mut2 plasmids are unstable (6 marks).

5. In the case of Mut1 how could you test your hypothesis (6 marks)?

Figure 1. Map of plasmids used to investigate the function of gene X in yeast. Threeplasmids 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 BIOLOGYAND 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


315 0.56


390 1.03



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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

(ii) the doubling time for the cell cultures at 37 C and 25 C

(iii) stating any assumptions, estimate the value of A600 for the 25 C sample taken at3 hours after induction of protein expression.

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)

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.


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Question 2 What conclusions do you draw from the SDS-PAGE results about the

form of the expressed protein at the two temperatures? (5 marks)

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 imidazolegradients (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)

Question 4 Outline how would you quantitatively assess the proportion of protein that

was produced in soluble form at the two temperatures? (6 marks)

problem specimen assembled

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