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Short questions – 1 point per question.

Q1 (1 point): Explain why a lettuce leaf wilts when it is placed in a concentrated salt solution.

Answer: Water is sucked out of the cells by osmosis (this reduces the turgor pressure on the cell

walls, which would otherwise keep the cells stiff).

Q2 (1 point): Put a cross by the correct answer(s) below.

The Na+/K+ pump carries out:

a. Primary active transport

b. Secondary active transport

c. Symport

d. ATP hydrolysis

Answer: a and d are correct.

Q3 (1 point): Name an enzyme that takes part in glycolysis, uses an “induced-fit” mechanism and

carries out substrate-level phosphorylation. NB there is only one correct answer.

Answer: Hexokinase.

Q4 (1 point): The following enzyme-catalysed reaction has reached equilibrium:

Glucose-6-phosphate Fructose -6-phosphate

What can be done to obtain a greater amount of fructose-6-phosphate? Mark the correct answer(s)

with a cross.

a. Add more enzymes to the reaction

b. Add ATP

c. Remove fructose-6-phosphate from the reaction mixture

d. Add more glucose-6-phosphate to the reaction mixture

Answer: c and d are correct.

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Q5 (1 point): Which coenzyme can take up electrons in redox reactions, contains an adenine base

and is required in glycolysis? NB there is only one correct answer.

Answer: NAD+

Q6 (1 point): Which enzyme can make a DNA copy using an RNA template? Mark the correct

answer(s) with a cross.

a. RNAse

b. DNA polymerase

c. Reverse transcriptase

d. DNA ligase

Answer: c

Q7 (1 point): A cDNA library cannot be used to isolate the promotor region of a gene. Why not?

Answer: cDNA only includes the transcribed regions of the gene, and therefore does not contain the

promoter region. A genomic library should be used instead.

Q8 (1 point): You have identified a gene that you wish to insert into a plasmid (shown below) and

introduce into E. coli cells.

Place the following steps in the correct order:

a. E. coli cells are plated out on Tetracycline-containing medium.

b. Treatment with DNA ligase.

c. The desired gene sequence and the plasmid are each treated with PstI.

d. E. coli cells are transformed with the DNA.

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e. The restriction fragments are mixed.

f. E. coli cells are plated out on Ampicillin-containing medium.

Answer: c, e, b, d, a, f

Q9 (1 point): Number the Carbon atoms in the ring form of ribose below.

Answer:

Q10 (1 point): What is ribose called, when the –OH group at position 2 is replaced with –H?

Answer: Deoxyribose.

Q11 (1 point): What is the name of the discontinuous sequences of DNA that are initially

synthesized at the lagging strand?

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Answer: Okazaki fragments.

Q12 (1 point): Which base can undergo deamination and thereby course mutations?

Answer: Cytosine.

Q13 (1 point): A linear DNA fragment is digested by a restriction enzyme that cuts at two different

positions. How many bands will be observed after gel electrophoresis of the digested DNA? How

many bands would you observe if a plasmid is cut two places?

Answer: Three and two bands, respectively.

Q14 (1 point): The following three pieces of sequences originate from the same DNA string and

have been found by sequencing. What is the sequence of the full DNA fragment?

5’ AGCGTTAG 3’

5’ CCGGTAAA 3’

5’ AGCCGGTA 3’

Svar: 5’ AGCGTTAGCCGGTAAA 3’

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Q15 (1 point): The figure below depicts a phase in a biological process that ultimately leads to cell

division. What is the biological process called and which phase is depicted?

Answer: Meiose, anaphase I (separation of the homologous chromosomes).

Q16 (1 point): In pea plants, the allele that encodes tall plants (H) dominates the allele that encodes

short plants (h). How large a fraction of the progeny will be tall in a monohybrid cross between two

plants that are both heterozygous with regards to height? (You may want to use a Punnett square to

figure this out).

Answer: ¾ of the progeny will be tall (1/2 will have the genotype Hh and ¼ will have the genotype

HH).

Q17 (1 point): In pea plants, the allele that encodes spherical seeds (S) dominates the allele that

encodes wrinkled seeds (s). You want to determine whether a pea plant with spherical seeds is

homozygous or heterozygous by performing a test cross with a plant with wrinkled seeds. How

large a fraction of the progeny do you expect to have spherical seeds and how large a fraction do

you expect to have wrinkled seeds, if the plant in the P generation with the spherical seeds is

homozygous, respectively heterozygous? (Note: Two answers are required – one if the plant in the

P generation with the spherical seeds is homozygous and one if it is heterozygous. You may want to

use a Punnett square to figure this out).

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Answer: If the plant with the spherical seeds is heterozygous (genotype = Ss) and it is crossed with

a plant with wrinkled seeds (genotype = ss) half the progeny will have spherical seeds (genotype =

Ss) and half will have wrinkled seeds (genotype = ss). If the plant with the spherical seeds is

homozygous (genotype = SS), all the progeny will have spherical seeds (genotype = Ss).

Q18 (1 point): This exercise is regarding a dihybrid cross, in which we look at the allele for tall

plants (H), which dominates the allele for short plants (h), and the allele for spherical seeds (S),

which dominates the allele for wrinkled seeds (s). The genes for plant height and seed shape are not

linked.

Which of the below genotypes would you not expect to find in the progeny of a cross between

plants with these two genotypes: HhSs x hhss.

A. hhss

B. HhSS

C. Hhss

D. hhSs

E. HhSs

Answer: You would not expect to see the genotype HhSS (B).

Q19 (1 point): This exercise is regarding red-green colour blindness in humans. A woman and a

man, both with normal colour vision, have a son, who is red-green colour-blind. Did the son inherit

the disease-causing allele from his mother or his father? Explain your reasoning.

Answer: The allele that causes red-green colour blindness is located to the X chromosome. Since

sons inherit a Y chromosome from their father and an X chromosome from their mother, the son

must have inherited the disease-causing allele from his mother. Since the disease-causing allele is

recessive, it is not expressed in the phenotype of the mother, who has two X chromosomes.

Q20 (1 point): Below, a family tree is shown for a family in which a hereditary disease occurs.

Squares represent men, while circles represent women. Black squares/circles represent people with

the disease, while white squares/circles represents healthy individuals. Since we are dealing with a

rare hereditary disease, it is safe to assume that individuals that are married into the family do not

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carry the disease-causing allele. The dashed line indicates that two cousins marry. Both of them are

healthy. They now expect their first child. What is the probability that this child will inherit the

disease? Explain your reasoning.

Answer: The disease-causing allele is not sex-linked and must be dominant, since it is characteristic

for this type of allele that the disease occurs in both men and women, that all diseased individuals

have at least one parent, who is also diseased, and that approximately half of the children of a

diseased and a healthy parent inherits the disease. Accordingly, the two cousins, who are both

healthy, do not carry the disease-causing allele (or they would themselves be sick). Hence there is

no risk of passing it on to their child, who has 0% risk of inheriting the disease.

Q21 (1 point): The TATA box is the area in promotors of eukaryotes, where the DNA string first

starts denaturing so that the transcription machinery can get access. The TATA box mainly consists

of A-T basepairs. Why is this an advantage, considering that the DNA string should denature?

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Answer: A-T basepairs are only held together by two hydrogen bonds, while C-G basepairs are held

together by three hydrogen bonds.

Long questions – from 4 to 9 points per question.

Q22 (6 points): The sketch below shows a polypeptide chain:

(i) Mark clearly on the sketch:

A: A peptide bond

B: The N-terminus of the polypeptide

C: The C-terminus of the polypeptide

D: A hydrophobic side chain

Answer: for completeness, all peptide bonds and hydrophobic side chains are shown below, even

though only one example of each was requested. See Ch. 3 p43-44.

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(ii) Can this polypeptide chain form a disulphide bridge? Explain why/ why not?

Answer: No, it contains no cysteines.

(iii) Write the amino acid sequence of the polypeptide in single-letter code (one letter

per amino acid)

Svar: MARDGDEL

(iv) Would this amino acid sequence be suitable as part of a membrane-spanning alpha

helix? Explain why/ why not?

Answer: It is highly unlikely. Membrane-spanning helices must interact with the hydrophobic

interior of the lipid bilayer and therefore typically contain many hydrophobic amino acid side

chains. The above sequence contains many charged side chains and only 3 hydrophobic amino acids

(M, A, L).

(v) Write an mRNA sequence that would be translated to the above amino acid

sequence (write only the coding part of the mRNA). NB there is more than one correct

answer.

 

Answer: One sequence that codes for the above amino acid sequence is:

5’ AUGGCUCGUGAUGGUGAUGAACUU 3’

But there are many other possibilities (see the genetic code on p.299).

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Q23 (4 points): The DNA fragment shown in Figure 1 is cleaved by the restriction enzym EcoRI as

indicated. The number in parenthesis shows the position of the cleavage site. The total length of the

DNA fragment is 4000 bp. Small parts of the DNA sequence is known as shown.

Figure 1: DNA fragment with a total length of 4000 bp.

(i) The figure below depicts a gel on which marker DNA of known size has been

run. Sketch the location of the bands that will appear, if the DNA fragment

shown in Figure 1 is cleaved by EcoRI and afterwards run on the gel along

with the marker DNA.

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Answer:

The gel is blotted onto a membrane and hybridized to the following radioactively labelled probe:

3’ CCCTCCCGTAGAGCGCTTAAAGCATTTCGCG 5’

After hybridization, the membrane is put on a x-ray film.

(ii) Which band will be apparent on the resulting picture? Explain your reasoning.

Answer: The band at 2500 bp will be apparent after the hybridization, since this DNA fragment

contains a sequence that is complementary to the radioactive probe.

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Q24 (9 points): The DNA string shown in Figure 2 originates from the Y chromosome. Note that it

contains the sequence CCTT (or AAGG on the complementary string) repeated several times.

Different individuals will have the sequence CCTT repeated a varying number of times (from 1 –

approximately 100 times), while all individuals will have the same flanking sequences (marked in

bold). The sequence does not encode protein.

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5’ TTACGAGCTTTGGGCTATGCCTCAGTTTAAAATACATGCCTGCCTTCCTTCCTTCCTTCCTTCCTTCCTT

3’ AATGCTCGAAACCCGATACGGAGTCAAATTTTATGTACGGACGGAAGGAAGGAAGGAAGGAAGGAAGGAA

140

CCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCC

GGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGG

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TTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTT

AAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAA

CCTTGAAAGAGTCTCTCTGTTACCCAGGCCC 3’

GGAACTTTCTCAGAGAGACAATGGGTCCGGG 5’

Figure 2: DNA sequence from the Y chromosome.

(i) Does the sequence shown in Figure 2 originate from a man or from a woman?

Answer: The sequence has to originate from a man, since women do not have a Y chromosome.

You want to amplify the sequence shown in Figure 2 using PCR. This will enable you to compare

the size of the PCR product that is produced when using DNA from a crime scene, with the size of

the PCR product that is produced when using DNA from suspects.

(ii) Design two primers that can be used for amplifying the area with the repeated

sequences of CCTT. The two primers should each consist of 20 nucleotides. Write

the sequence of the two primers and remember to mark the 5’ and 3’ ends. Also

mark where the two primers will bind on the sequence in Figure 2.

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Answer: 70

5’ TTACGAGCTTTGGGCTATGCCTCAGTTTAAAATACATGCCTGCCTTCCTTCCTTCCTTCCTTCCTTCCTT

3’ AATGCTCGAAACCCGATACGGAGTCAAATTTTATGTACGGACGGAAGGAAGGAAGGAAGGAAGGAAGGAA

A: 5’ TGGGCTATGCCTCAGTTTAA 3’

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CCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCC

GGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGG

210

TTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTTCCTT

AAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAAGGAA

3’ TCAGAGAGACAATGGGTCCG 5’

CCTTGAAAGAGTCTCTCTGTTACCCAGGCCC 3’

GGAACTTTCTCAGAGAGACAATGGGTCCGGG 5’

Primer A: 5’ TGGGCTATGCCTCAGTTTAA 3’

Primer B: 3’ TCAGAGAGACAATGGGTCCG 5’

(iii) List two reasons why you should design your primers so that they only bind the

flanking sequences and not the area with the repeated CCTT sequences.

Answer: Firstly, only the flanking sequences are present in all men. No PCR product will be

produced, if the primers are designed so that they bind the area with the repeated CCTT sequences

and you then use DNA from a man who only have one CCTT repeat. Secondly, the primers should

be designed so that they are specific, meaning that they each only bind in one place. Otherwise, you

will get a PCR product that consists of DNA strings of varying lengths, or possibly no PCR product

at all.

(iv) Besides DNA from a biological sample and primers, which ingredients are needed

for PCR?

Answer: Besides DNA and primes, the four nucleotides (dATP, dGTP, dCTP, dTTP) and a heat

stable DNA polymerase is needed (and salts and buffer to keep the pH stable).

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(v) What is the size of the PCR product, if you use the primers that you designed in (ii)

and the DNA sequence that is shown in Figure 2?

Answer: 229 bp, if you have chosen the same primers as shown in the answer of (ii). But there are

multiple correct answers. If you have chosen your primers so that they only bind the flanking

sequences, the PCR product will be from 172 - 241 bp.

A count has been murdered. DNA from a hair that was found on the crime scene has been purified

(the count himself was bald, so it cannot be from him). Using this DNA and a set of primers binding

the flanking regions of the DNA sequence shown in Figure 2, PCR is performed. DNA is also

purified from the three suspects: The count’s brother, the butler and the gardener. PCR is performed

using this DNA and the same set of primers. The products of all four PCR reactions are run on a

gel, as depicted in the figure below.

Here, the content of the five lanes of the gel is listed:

A: Marker DNA containing DNA fragments of known length.

B: Product of PCR using DNA purified from the hair from the crime scene.

C: Product of PCR using DNA purified from the count’s brother.

D: Product of PCR using DNA purified from the butler.

E: Product of PCR using DNA purified from the gardener.

(vi) Who is it most probable that the hair belongs to? Explain your reasoning.

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Answer: It is most probable that the hair belongs to the butler, since DNA from the hair and from

the butler results in a PCR product of equal size.