Genetic code tutorial sheets

CHROMOSOMES and THE GENETIC CODE

1. The diagram below shows a portion of a DNA molecule. Show with the help of diagrams how this portion replicates.

(4 marks)

2. Which of the following steps in DNA replication involves the formation of new phosphodiester bonds?

a) Initiation of an origin of replication

b) Elongation by a DNA polymerase

c) Unwinding of the double helix

d) Termination

3 . The difference in leading- versus lagging-strand synthesis is a consequence of –

a) the antiparallel configuration of DNA

b) DNA polymerase III synthesizing only in the 5’-to-3’ direction

c) The activity of DNA ligase

d) Both (a) and (b)

4. Okazaki fragments are—

a) synthesized in the 3’-to-5’ direction

b) found in the lagging strand

c) found in the leading strand

d) made of RNA

5. The replication of a DNA molecule occurs in a series of coordinated steps requiring the action of several enzymes.

a) What is DNA replication? (1 mark)

The process by which new copies of DNA are produced.

b) DNA replication is termed semi-conservative replication. What does this imply about the process? (1 mark)

The newly formed DNA consists of an original strand and a synthesised one.

c) One of the first steps in DNA replication involves the formation of a replication fork. Name the enzyme responsible for the formation of a replication fork and briefly describe how this enzyme modifies a double helix to form this structure. (1 mark)

Helicase is the enzyme. It breaks the hydrogen bonds between the complementary bases. 

d) What are the roles of DNA polymerase in the replication process? (2 marks)

Catalysis the polymerisation of deoxyribonucleotides alongside a DNA strand which it “reads” and uses as a template.

Plays a role in proofreading.

Removes RNA primers and replaces them with DNA.

e) The lagging strand of the replication fork is synthesized in short fragments known as Okazaki Fragments. Suggest a reason for the synthesis of these short fragments. (2 marks)

Since DNA polymerase III can lay down new nucleotides only in the 5’ to the 3’ direction.

f) What is the role of ligase in the formation of the lagging strand? (1 mark)

Ligase joins the Okazaki fragments together by helping the formation of phosphodiester bonds.

g) During which part of the cell cycle is DNA replication likely to occur? (1 mark)

During S phase of interphase.

THE GENETIC CODE6. The following are mRNA codons for three

amino acids and a stop codon.Glutamine GAA Serine UCA Lysine AAA Stop

codonUAA

GAG UCG AAG

a) Give an appropriate nucleotide base sequence for:

i) the DNA strand complementary to an mRNA codon for serine; (1 mark)mRNA codon = UCA and UCG DNA codon = AGT and AGC, respectively

a) Give an appropriate nucleotide base sequence for:  

ii) an anticodon for glutamine. (1 mark)

mRNA codon = GAA and GAG

anticodon = CUU and CUC, respectively

b) Giving examples from the codons above, explain why the third base of a codon is less critical than the first two. (2 marks)

The first two bases determine the amino acid e.g. GAA and GAG both code for glutamine.

Glutamine GAA Serine UCA Lysine AAA Stop codon

UAA

GAG UCG AAG

7. The flow diagram shows the course of an experiment with a virus and the bacterium Escherichia coli.

a) Which molecule in the virus was being labelled with:

i) radioactive sulphur, 35S (1 mark)

Amino acids.

ii) radioactive phosphorus, 32P?

(1 mark)

Phosphate group of nucleotides.

b) Explain how the viruses collected from culture A became labelled. (1 mark)

Viruses use labelled amino acids to build their capsid.

c) Explain the difference in radioactivity of the viruses at the end of the experiment.

(2 marks)

Radioactive sulphur: No radioactive viruses were collected as the protein coat did not enter the bacteria.

Radioactive phosphorus: Some radioactive viruses were collected as they injected their DNA into bacteria.

d) Suggest why 14C was not used to label the viruses in this experiment. (1 mark)

As carbon is present in both amino acids and nucleotides. It would not be possible to trace proteins and nucleic acids if radioactive carbon is used as this is present in both.

e) This classical experiment was carried out by Hershey and Chase in1952. What important conclusion did they come to, regarding phage genetic material, following the results of this experiment? (1 mark)

DNA codes for the protein coat.

8. Cells of the bacterium E. coli were grown for many

generations on a medium containing the heavy isotope of

nitrogen, 15N. This labelled the entire DNA in the bacteria.

The cells were transferred to a medium containing 14N and

allowed to grow. During each generation of bacteria, the DNA

replicates once. Samples of the bacteria were removed from

the culture after one generation time and after two generation

times. The DNA from each sample was extracted and

centrifuged. As the DNA containing 15N is slightly heavier

than that containing 14N, the relative amounts of DNA labelled

with 14N and 15N can be determined.

The diagram below shows two reference tubes and the

results of this experiment.

a) Which part of the DNA molecule in the original culture would have been labelled with 15N? (1 mark)

Nitrogenous base.

b) Explain why the DNA occupies an intermediate position after one generation in the 14N containing medium.

Weight of DNA is intermediate between heavy and light types. It is made of one heavy and one light strand. (2 marks)

c) Complete the diagram to show the position of the band or bands of DNA after two generations in the 14N-containing medium (Tube X). (1 mark)

d) This neat experiment was designed by Matthew Meselson and Franklin Stahl in 1958 in order to evaluated 3 then-current models of DNA replication. Which of these models for DNA replication is supported by the results of this experiment and is still supported today? (1 mark)

Semi-conservative replication.

PROTEIN SYNTHESIS9 Read through the following account of DNA and protein

synthesis, and then write on the dotted lines the most appropriate word or words to complete the account.

A molecule of DNA (deoxyribonucleic acid) consists of two polynucleotide strands with complementary pairs of organic bases. Each nucleotide consists of a sugar called DEOXYRIBOSE, a PHOSPHATE group and an organic base. Pairs of bases are held together by weak HYDROGEN bonds. During the first stage of protein synthesis the DNA strands are separated and a complementary molecule of mRNA is synthesised in the nucleus using one strand of DNA as a TEMPLATE .

Where the base cytosine occurs in the DNA strand, the newly synthesised molecule contains the base GUANINE and where the DNA has adenine, the newly synthesised molecule has URACIL. The newly synthesised molecule then passes into the cytoplasm and becomes attached to a RIBOSOME where synthesis of the polypeptide chain occurs.

10 a) Complete the diagram below by adding the corresponding nucleotide bases to:

i) the DNA strand;

ii) the tRNA molecules to form the appropriate anticodons for the amino acids X, Y and Z.

b) The following are mRNA codons for three amino acids:

tryptophan UGG lysine AAG isoleucine AUU AAA AUC AUA

i) Identify amino acid Z. (1 mark) Isoleucine.

tryptophan UGG lysine AAG isoleucine AUU AAA AUC AUA

ii) Which of the amino acids X, Y and Z has or have a degenerate code? Explain your answer. (2 marks)X and Z – different codons code the same amino acid.

Lysine Isoleucine

Tryptophan

c) What is the function of the base sequence of mRNA? (3 marks)

It determines the sequence of amino acids in a polypeptide. Every three bases make up a codon and code for one amino acid. A start codon (AUG) denotes where the polypeptide formation starts while a stop codon (UGA, UAA and UAG)

 d) Describe the part played by ribosomes in protein synthesis. (2 marks)

The ribosome binds mRNA and tRNA molecules. rRNA in the large subunit acts as an enzyme that links amino acids together as peptide bonds form.

e) Describe the role of transfer RNA in protein synthesis. (3 marks)

A tRNA molecule binds an amino acid in the cytoplasm and delivers it to the ribosome. Its anticodon is complementary to the mRNA codon.

11 a) A piece of mRNA is 660 nucleotides long but the DNA non-template strand from which it was transcribed is 870 nucleotides long.

i) Explain this difference in the number of nucleotides. (1 mark)

DNA is longer due to introns. These were removed from mRNA.

11 a) A piece of mRNA is 660 nucleotides long but the DNA template strand from which it was transcribed is 870 nucleotides long.

ii) What is the maximum number of amino acids in the protein translated from this piece of mRNA? Explain your answer.

(2 marks)

Due to the triplet code: 660/3 = 220

Max No. of amino acids: 220-1 (due to a stop codon that does not code for an amino acid)

= 219

b) Complete the table to give two differences between the structure of mRNA and the structure of tRNA.

mRNA. tRNA.1. Has codons2. Linear molecule 3. Larger

AnticodonsMolecule is foldedSmaller

WORK OUT Nos. 18, 19 [about lac operon] for

Friday 7 March 2014

CONTROL OF GENE EXPRESSION IN PROKARYOTES

14 Control of gene expression can occur at which of the following steps?

a) Splicing of pre-mRNA into mature mRNA

b) Initiation of translation

c) Initiation of transcription

d) All of the above

15 Negative control of transcription in a prokaryotic cell involves __________molecules that alter the conformation of _______________ proteins that bind to DNA and prevent transcription.

e) operator; repressor

f) activator; RNA polymerase

g) activator; operator

h) effector; repressor.

16 What is an operon?

a) A region of DNA involved in regulation of transcription

b) A cluster of genes that are expressed as a single unit

c) A DNA-binding motif

d) A regulatory protein that enhances transcription

17 What effect would the presence of lactose have on a lac operon?

e) The repressor would bind to the operator site of the operon

f) Lactose will bind to the operator site of the operon

g) The lac operon would be transcribed.

h) It would have no effect.

18 The figure below shows a diagrammatic representation of the lac operon including the Promoter region, Operator region and three structural genes.

a) What is an operon? (2 marks)

An operon is a segment of DNA containing adjacent genes including structural genes, an operator gene, and a regulatory gene. It is a functional unit of transcription and genetic regulation.

b) Briefly describe the function of:

i) The promoter region; (2 marks)

The DNA region, usually upstream to the coding sequence of a gene or operon, which binds and directs RNA polymerase to the correct transcriptional start site and thus permits the initiation of transcription.

ii) The operator region; (2 marks)

A gene that activates the production of messenger RNA [transcription] using adjacent structural genes. The repressor protein binds to the operator region.

iii) The three structural genes lacZ, lacY and lacA;(2 marks)

These are three structural genes coding for three different proteins – beta galactosidase, galactosidase permease and galactosidase transacetylase.

c) List ONE way in which gene expression in prokaryotes differs from gene expression in most eukaryotes.

Eukaryotes have a nuclear envelope, which prevents simultaneous transcription and translation.

Gene expression involves a polycistronic mRNA but a monocistronic one in eukaryotes.

(1 mark)

19. Colonies of the bacterium Escherichia coli grow rapidly on culture media based on glucose. When transferred to a culture medium based on lactose, growth rates of the colonies slow down temporarily but subsequently reach the same rates characteristic of glucose-based media. Analysis of the colonies growing on lactose-based media indicated the presence of –galactosidase, an enzyme that hydrolyses lactose to glucose. This enzyme is not present in colonies grown on glucose-based media.

a) –galactosidase is an inducible enzyme. What is an inducible enzyme? (2)

Produced only in the presence of the inducer molecule, i.e. lactose.

b) What is the principal difference between inducible enzymes and constitutive enzymes? (2)

Inducible enzymes are made only in the presence of their inducer molecules, while constitutive enzymes are made all the time.

c) Briefly explain why –galactosidase is not synthesised when E. coli is cultured on media lacking lactose. (5)

Repressor molecules are made all the time in bacteria.

In the absence of lactose, the repressor molecule is active and binds to the operator region on the DNA.

The binding of the repressor to the DNA, prevents RNA polymerase from binding to the promoter region.

This means that transcription cannot occur. If mRNA is absent, then no translation occurs and enzymes are not made.

d) Some E. coli will synthesise –galactosidase in the absence of lactose. Suggest an explanation for this observation. (2)

A mutation in the gene results in a different shape of the repressor molecule which cannot bind to the operator , thus RNA polymerase can bind to its promoter and transcription proceed. Translation produces –galactosidase.