Homeostasis and Cell signalling

Multiple Choice Questions1Viral genomes vary greatly in size ranging from four genes to several hundred genes. Which of the following viral features is most apt to correlate with the size of the genome?

Asize of the viral capsomeres

BRNA versus DNA genome

Cdouble stranded versus single stranded genomes

Dsize and shape of the capsid

2Emerging viruses arise by

Amutation of existing viruses.

Bthe spread of existing viruses to new host species.

Cthe spread of existing viruses more widely within their host species.

Dall of the above.

3To cause a human pandemic, the H5N1 avian flu virus would have to

Abecome capable of human to human transmission.

Bdevelop into a virus with a different host range.

Cbecome much more pathogenic.

Dundergo an antigenic shift.

4A bacterium is infected with an experimentally constructed bacteriophage composed of T2 phage protein coat and T4 phage DNA. The new phages produced would have

AT2 protein and T2 DNA

BT2 protein and T4 DNA

CT4 protein and T2 DNA

DT4 protein and T4 DNA

5Some statements about the lytic cycle of a T4 bacteriophage are listed below.

1 Bacteriophage tail fibres attach to receptors on the bacterial cell membrane. 2 Bacteriophage enzymes break down bacterial DNA. 3 Bacteriophage nucleic acid, enzymes and other proteins are injected into the bacterial cell. 4 New bacteriophages assembled. 5 Bacteriophage DNA replicates and codes for new bacteriophage proteins. 6 Release of bacteriophages.

Which sequence of these statements correctly describes the lytic cycle?

A1, 2, 3, 4, 5, 6

B1, 3, 2, 4, 5, 6

C1, 3, 2, 5, 4, 6

D1, 3, 5, 2, 4, 6

6The graph shows the changes in the core protein (capsid) concentration of the human immunodeficiency virus (HIV) in the blood stream after an HIV infection. The protein is detected using antibodies.

What explains the apparent absence of the protein between three months and seven years?

AThe core protein attaches to a CD4 receptor, therefore, it is not found in plasma.

BThe lipid bilayer envelope masks the core protein.

CThe patients own antibodies destroy the core protein.

DThe virus enters the latent phase.

7A mutated lacI gene resulting in the failure of allolactose binding to the repressor protein would cause

Areduced expression of operon.

Bconstitutive expression of the operon.

Creversible binding of lac repressor protein to operon when lactose is present.

Dpermanent binding of lac repressor protein to operon when lactose is present.

8In generalised transduction, defective virus are formed as a result of

Aviral enzymes cutting the host DNA such that the host DNA is assembled into the new virus.

Bproduction of host enzymes by virus which nicks its own DNA such that it can be assembled into the new virus.

C no shut down of host DNA production such that host DNA or the virus DNA can either be assembled into the new virus.

D integrate of virus DNA into host DNA and during excision the viral genome carries along with it the host DNA to be assembled into the new virus.

9The diagram shows the lac operon.

PILacI

PlacOLacZLacYLacA

Mutation in which of the following DNA sequence(s) would result in the constitutive synthesis of -galactosidase?

ALacZ and LacY

BLacI and Plac

CO and LacI

DPI and Plac

10Which component/s make up a virus?

Aprotein particles only

BDNA or RNA and a protein coat

CDNA and RNA and a protein coat

DDNA in a nucleus, RNA, ribosomes, plasma membrane, and a cell wall

11Two strains of the bacterium Streptococcus pneumoniae were used in an experiment on mice. The smooth (S) strain is a virulent strain whereas the rough (R) strain is a non- virulent strain. The diagram below shows the results of the experiment.

Which of the following accounts for the living S cells detected in the mouse blood sample, even though the original injection mixture contained only heat-killed S cells and living R cells?

ABinary fission

BConjugation

CTransduction

DTransformation

12A microbiologist analyzed the DNA of two E. coli cells, an F+ cell and an F- cell, before and immediately after their conjugation. He found that ________________________________.

Aboth cells lost some genes and gained others

Bboth cells gained genes but lost none of their original genes

Cone cell lost genes and the other gained genes

Done cell gained genes and the genes of the other were unchanged

13How does a bacterial chromosome differ from a eukaryotic chromosome?

Bacterial chromosomeEukaryotic chromosome

Aa circular, single-stranded DNA moleculea linear, single-stranded DNA molecule with many associated proteins

Ba circular, double-stranded DNA molecule with associated proteinsa linear, double-stranded DNA molecule with many associated proteins, including histones

Ca circular, double-stranded DNA molecule with associated histone proteinsa linear, double-stranded DNA molecule with many associated proteins

Da linear, double-stranded DNA molecule with many associated proteinsa circular, double-stranded DNA molecule with very few attached protein molecules

14Which does not occur during the formation of bacterial messenger RNA?

Aformation of hydrogen bonds

Binitiation of translation

Closs of water molecules

Dpolymerization of deoxyribonucleotides

15Which features of viruses account for them being obligate parasites?

1. All viruses are very small, ranging in size from 20-300nm.2. Each virus contains only one type of nucleic acid.3. Viruses can be crystallised.4. Viruses cannot synthesise ATP.5. Viruses have no cellular structure.6. Viruses have no enzymes involved in metabolism outside a host cell.

A1, 3 and 5

B1, 4 and 6

C2, 3 and 5

D2, 4 and 6

Anglo-Chinese Junior CollegeRevision Package H2 Genetics of Viruses & Bacteria

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16Some events that take place during the reproductive cycle of influenza virus are listed.

1. The virus enters the cytoplasm surrounded by a coated vesicle.2. The nucleocapsid is released and viral RNA replicated by RNA polymerase.3. Each replicated viral RNA is given a new envelope.4. Viral hemagglutinin proteins bind to receptors in the host cell surface membrane.5. Hemagglutinin and neuraminidase are produced by the host cells ribosomes.6. The replicated viral RNA is transcribed to mRNA.

Which sequence of events correctly describes the reproductive cycle of the virus?

A1 2 3 6 5 4

B1 2 6 5 4 3

C4 2 5 1 6 3

D4 1 2 6 5 3

17The synthesis of the amino acid tryptophan is E. coli is controlled by the trp operon. The amino acid tryptophan combines with an inactive trp repressor protein to activate it. This trp repressor protein binds to the operator controlling the expression of structural genes for the production of tryptophan. What does this system ensure?

ACells are able to absorb tryptophan.

BTryptophan is constantly manufactured.

CTryptophan is manufactured only when levels are low.

DTryptophan inhibits translation at the ribosome.

18Which statement CORRECTLY describes the control of transcription of the genes involved in the breakdown of lactose in Escherichia coli?

AA repressor protein binds to the operator and the genes are switched on.

BA repressor protein binds to the operator and the genes are switched off.

CA transcription factor binds to the promoter and the genes are switched on.

DA transcription factor binds to the promoter and the genes are switched off.

19Which drug would be most effective in treating viral infections?

AOne that competes with the virus for the same binding site on the host receptor.

BOne that induces the body to produce antibodies.

COne that inhibits the action of viral ribosomes.

DOne that interferes with the synthesis of viral nucleic acids.

20Regarding prokaryotic reproduction, which statement is correct?

AProkaryotes form gametes by meiosis.

BProkaryotes feature the fusion of haploid gametes, as do eukaryotes.

CProkaryotes exchange some of their genes by conjugation, the fusion of haploid gametes, and transduction.

DMutation is a primary source of variation in prokaryote populations.

Structured Questions

1. (a) Briefly describe the structural components of viruses. [1]

HIV enters blood cells via CCR5 receptor molecules in the cell surface membranes. Fig. 1.1 shows a CCR5 receptor, which is a protein.

Fig.1.1

Source: Samson M, Libert F et al. Nature 382 (6593) p722 1996(b) Using information from Fig. 1.1, describe the protein structure of the CCR5 receptor. [3]

(c) Some people carry a mutant form of a gene which results in a different form of the CCR5 receptor. Suggest how this may result in protection from HIV. [1]

(d) Four possible combinations for the binding of catabolite activator protein (CAP) and Lac repressor are shown in Fig. 1.2.

For each of the four combinations, indicate on the left-hand side of Fig. 1.2 which sugar(s) must be present (using + ) or absent (using - ), and on the right hand side whether the operon is expected to be turned On or Off .

Fig.1.2

(e) Bacterial cells can take up the amino acid tryptophan from their surroundings. If the external supply is insufficient, they can synthesize tryptophan from small molecules in the cell. Tryptophan repressor inhibits transcription of the genes in trp operon. Upon binding tryptophan, the tryptophan repressor binds to a site in the promoter of the operon. Why is such tryptophan-dependent binding to the operon a useful property for tryptophan repressor? How does it illustrate that the trp operon is a repressible system? [2]

(f) Fig. 1.3 shows how a trp-lac fusion operon is constructed.

Explain the condition(s) required for beta galactosidase to be expressed in bacteria cells that have been transformed with the trp-lac fusion operon.[2]

[Total: 10]

2 (a) In a study to examine the effectiveness of bacteriophages in treating colibacillosis, a fatal disease caused by E. coli in poultry, broiler chickens were first subjected to an aerosol spray containing bacteriophages on day 0. They were then separated into five treatment groups. Each treatment group was subsequently injected with E. coli on days 0, 1, 2, 3 and 4 respectively. The mortality rate of each group was determined after 21 days. The result of the study is represented by Fig. 2.1 below.

Fig. 2.1

Fig. 2.1(i) With reference to Fig. 2.1, compare the general trends observed in the control groups and the groups that have been treated with bacteriophages. [3]

(ii) Discuss ways in which the farmer can make use of the results in Fig. 2.1 to treat colibacillosis in his poultry. [2]

(bi) Suggest why the use of bacteriophages is a better possible alternative to antibiotic therapy for the chickens. [1]

(b) (i) A person infected with the H1N1 influenza virus does not develop the symptoms. Physical examination shows that the persons immune system has produced an antibody that binds to the haemagglutinin molecules. Suggest why the person does not develop influenza-like symptoms. [2]

(ii) After a few days, the person begins to develop influenza symptoms. Tests for the antibody mentioned in (b) (i) show that it is still present in the blood. Account for this observation and comment on the implication of this observation for developing a H1N1 influenza vaccine. [2]

[Total : 10]

3Fig. 3.1 shows the percentage of regulatory sites spanning the indicated base positions in operons found in E. coli. Activator binding sites% of regulatory sites at indicated positions in different operonsBase position

Repressor binding sites

Fig. 3.1

(a)With reference to Fig. 3.1,

(i) indicate using on the x-axis of the graph the base position of the promoter. [1]

(ii)outline the role of the repressor binding site in an operon. [1]

(iii) state and explain whether the structure of the lac operon conforms to the data shown. [3]

[Total: 5]4Fig. 4.1 shows the structure of the human immunodeficiency virus, HIV.

Fig. 4.1

(a) Name the parts A to C. [3]

A: ____________________________B: _____________________________C: _____________________________

HIV instructs the T-lymphocytes to reproduce more viruses. During this process, the cell replicates viral DNA and produces viral proteins, which will be used to make new viral particles. These HIV particles will be released from the infected T-lymphocyte through budding, so that more T-lymphocytes will be infected. Fig. 4.2 is an electron micrograph showing the process of HIV particles leaving a T-lymphocyte.

Fig. 4.2(b) Suggest why an infected T-lymphocyte that is producing HIV particles has a higher demand for amino acids than an uninfected lymphocyte. [1]

(c) Describe how HIV acquires the outer envelope. [3]

(d) Upon the release of HIV from an infected T-lymphocyte, HIV protease completes the maturation of the viruses by cutting viral polyproteins to form the HIV structural proteins and enzymes of the infective virus.

HIV protease inhibitors may be used to treat HIV patients.HIV protease is composed of two identical protein chains which forms a tunnel that wraps around the viral polyprotein chains, closing and holding it while it is hydrolysed to form the hiv structural proteins and enzymes.

Fig. 4.3 shows a molecular model of HIV protease with the protease inhibitor (shaded) attached.Fig. 4.3

(i) Explain how HIV protease inhibitors work in the treatment of HIV infection. [3]

(ii) Describe the structural properties of this inhibitor required for its action. [2]

[Total: 12]

5In 1951, Joshua Lederberg and Norton Zinder were testing for genetic recombination in the bacterium Salmonella typhimurium. The researchers used two different strains of bacteria: one which only grew when supplied with tryptophan and tyrosine, and a second strain which only grew when supplied with methionine and histidine. When either strain was plated on a minimal medium, no wild-type cells were observed. However, after the two strains were mixed, wild-type cells appeared at a frequency of about 1 in 105.

(a) Explain how genetic recombination typically occurs in eukaryotes. [2]

(b) Describe three methods by which genetic recombination may occur in bacteria. [3]

In a further U-tube experiment, the different strains of bacteria were placed in two arms of a U-tube, separated by a filter that prevents cell contact. The result of this experiment shows that recombinant strains could also be obtained. By varying the size of the pores in the filter, they found that the agent responsible for recombination was about the size of the virus P22, a known temperate phage of Salmonella.

(c) Explain what it means by temperate phage. [1]

(d) Outline the process by which a lytic bacteriophage leads to genetic recombination in bacteria. [4]

[Total: 10]

Essays

1(a)Discuss whether viruses are living or non-living organisms and explain why viruses are obligate parasites. [9]

(b)Some bacteriophages have both a lysogenic and a lytic cycle. Suggest reasons why it may be advantageous for a bacteriophage to have a lysogenic cycle. [4]

2(a)Distinguish between the reproductive cycles of lambda phage and HIV. [5]

(b)Explain how HIV infection causes diseases in humans. [9]

3(a)Discuss the importance of operons in bacteria. [4]