Microsoft Word - 2014 YR5 Hol Homework Questions

Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr 5 2014

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Question 1 (DNA and Genomics)

(a) The flow of information in the cell, sometimes called the Central Dogma of Molecular Biology, primarily involves two classes of macromolecules.

(i) State the two classes of macromolecules and their cellular location in eukaryotes. [2]

(ii) Use a simple flow chart to represent the Central Dogma in the space provided below. Include in the flow chart the processes by which information flows. [3]

(b) Table 1.1 below shows the relative proportions of the bases adenine, thymine, guanine and cytosine in DNA from different organisms.

Table 1.1

Explain the importance of the ratios of A to T and G to C to the structure of DNA. [2]

(c) During protein synthesis, tRNA and ribosomes are involved in the formation of a polypeptide chain.

(i) State the exact location in the cell where ribosomal subunits are made. [1]

(ii) Describe how the correct amino acid is attached to each tRNA molecule. [3]

[Total: 11 marks]


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Question 2 (DNA and Genomics)

a) (i) Describe the structural features of DNA. [2]

(ii) Outline the main features of DNA replication. [5]

b)

Table 2.2 The genetic code Use the genetic code in Table 2.2 as shown above to complete the table below. The columns represent transcriptional and translational alignments. Write down the directions in terms of 5 and 3 for the nucleic acids.

5 C DNA double

helix

G G G

C A

G

mRNA transcribed

G C A Appropriate

tRNA anticodon

Trp

Amino acids incorporated into

polypeptide [4]

[Total: 11 marks]


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Question 3 (DNA and Genomics) a) The main function of a chromosome is to act as a template for the synthesis of RNA

molecules, since only in this way does the genetic information stored in chromosomes become directly useful to the cell. RNA synthesis is a highly selective process.

Explain what is meant by the term template. [3]

b) Figures 3.1 shows transcription.

(i) Identify structures A to C and describe their function. [1]

(ii) Describe the function of RNA polymerase. [3]

(iii) State three differences between transcription and DNA replication. [3] [Total: 10 marks]

Figure 3.1 Transcription


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Question 4 (Bacteria)

a) An experiment was conducted to determine the identity of Substance X and Substance Y. Both substances are known to have an effect on the expression of -galactosidase in Escherichia coli. Substance X was added after 10 minutes, Substance Y was added after 20 minutes and both substances X and Y were added after 30 minutes. The results are shown in Fig. 5.1.

Fig. 5.1 a) (i) Suggest an identity for Substance X and Substance Y

Substance X:

Substance Y:

(ii) With reference to Fig. 5.1, explain how the expression levels of -galactosidase are affected by Substance X and Substance Y between 10 minutes to 40 minutes. [5]

Substance X added

Substance Y added

Substance X and

Substance Y added

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

0.0 10.0 20.0 30.0 40.0 50.0 60.0

Amo

un

t of

-ga

lact

osid

ase

(m

g)

Time (min)


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b) (i) Draw a simple diagram to show all the elements of the trp operon. [1]

(ii) Explain why it is useful for a bacterial cell to decrease expression of the trp genes when tryptophan is present. [2]

Table 5.1 below indicates the activity levels of the functional enzymes E, D, C, B and A in wild type bacterial cells in the presence and absence of tryptophan (Trp).

Activity level of enzymes/units

Enzyme Trp absent Trp present E 700 0 D 700 0 C 700 0 B 700 0 A 700 0

Table. 5.1

You have managed to obtain several bacterial mutants. Each mutant is the result of a single base-pair substitution in a single component of the trp operon. The activity level of functional enzymes E, D, C, B and A in the bacterial cells having these individual mutations is shown in Table 5.2.

Activity level of enzymes/units Mutant 1 Mutant 2 Mutant 3

Enzymes Trp absent

Trp present

Trp absent

Trp present

Trp absent

Trp present

E 700 700 700 0 0 0 D 700 700 0 0 0 0 C 700 700 700 0 0 0 B 700 700 700 0 0 0 A 700 700 700 0 0 0

Table. 5.2

c) (i) Explain why it is useful for a bacterial cell to decrease expression of the trp genes when tryptophan is present. [2]

(ii) A loss-of-function mutation in which component of the trp operon could explain the phenotype of mutant 3? Explain your choice. [2]

c) If the phenotype of mutant 3 is caused by a mutation in the trpR gene, explain how this mutation would affect the structure and function of the repressor protein. [3]

[Total: 15 marks]


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Question 5 (Bacteria) a) Figure 7.1 shows an electron micrograph of an E.coli (strain O104:H7) that can enter humans

through ingestion of contaminated food and cause food poisoning in humans. A serious outbreak of foodborne illness in Germany in May 2011 was caused by this novel strain of E.coli that had acquired the genes to produce Shiga toxins.

Figure 7.1 (i) State two structural differences between the chromosome of this type of microorganism

and a human cell. [2]

(ii) Explain how this bacteria strain of E.coli O104:H7 could have transferred the genes coding for Shiga toxins to normal E.coli bacteria in human gut. [4]

b) Figure 7.2 shows a diauxic growth curve of E. coli cultured in a medium with both glucose and lactose. Growth (!) of bacteria is measured in terms of optical density (OD) at wavelength of 650nm, concentration of lactose (!) and glucose (") are measured in millimolar (mM).

Figure 7.2 With reference to Figure 7.2, explain the bacteria growth rate (i) for the first 6 hours; [3]

(ii) from 6 to 8 hours; [2]

(iii) from 8 hours onwards. [5] [Total: 16 marks]


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Question 6 (Viruses)

a) Explain why viruses are necessarily obligate parasites. [3]

b) In 1911, Peyton Rous isolated a virus from chickens that rapidly produced tumours when injected into healthy birds. The virus was then named the Rous sarcoma virus (RSV). The life cycle of the virus is shown below in Figure 8.1.

(i) With reference Figure 8.1, describe the common features shared by RSV and retroviruses e.g. HIV. [3]

(ii) Describe two differences in the mode of entry of influenza virus and HIV. [4]

[Total: 10 marks]


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Question 7 (Pro and Euk)

a) Describe 2 roles of telomeres. [4]

b) Telomeres can be extended by an enzyme, telomerase. Telomerase comprises an RNA component which can be used as a template in the synthesis of telomeric DNA. Figure 4.1 illustrates the action of telomerase.

Figure 4.1

(i) Suggest how a synthetic single-stranded RNA molecule (not shown in Fig. 4.1) can be used to inhibit telomerase activity. [2]

(ii) Suggest how a mutation in DNA can result in increased telomerase activity in a cell. [2]

c) Figure 4.2 below shows the effect of histone acetylation/deacetylation on a piece of chromosome.

Figure 4.2

(i) With reference to Figure 4.2, explain the effect of histone deacetylation on gene expression. [3]

(ii) State and explain another process which can increase the effect of histone deacetylation on gene expression. [2]

[Total: 9 marks]


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Question 8 (Pro and Euk)

Using the table below, compare how gene regulation at translational level differs between a bacterium and a eukaryotic cell. Points of comparison for

translational control Prokaryote Eukaryote

Protecting the RNA and thus the stability of the mRNA

Recognition site for ribosomes

Location of binding for repressor

[Total: 8 marks]

Question 9 (Cell Division)

(a) A root was cut into ten transverse sections at different distances from the tip. The sections were stained and viewed under the microscope. The number of cells in mitosis were counted in each section and the results were used to determine the mitotic index.

This is calculated as follows:

Fig. 9.1 shows the mitotic index for the ten sections.


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(i) Using the information in Fig. 9.1, describe how the mitotic index changes along the length of the root.

[2]

(ii) Explain how the events in the mitotic cell cycle ensure that all the cells in the root are genetically identical.

[3]


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(b) The diagram shows the main stages in the formation of sperms in a human testis.

(i) Describe two ways, other than size, in which cells at anaphase of division A would differ from cells at anaphase in division B.

[2]

(ii) Explain why meiosis is important for sexual reproduction [3]

(iii) This man has haemophilia. Haemophilia is a sex-linked condition. Explain how a child that results from the zygote formed when sperm P fertilises an ovum may not inherit the haemophilia allele from its father.

[2]

[Total: 9 marks]


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Question 10 (Cell division)

Fig. 10.1 shows drawings of a cell at various stages in the mitotic cell cycle.

(a) List the letters shown in Fig. 10.1 in the order in which these stages occur during a mitotic cell cycle. The first stage has been entered for you. [1]

(b) Explain what is happening in stage D in Fig.10.1. [2]

(c) Describe and outline what happens to the DNA in the nucleus during stage A in Fig. 10.1. [3]

(d) State the importance of mitosis in the growth of a multicellular organism, such as a flowering plant or a mammal. [1]

[Total: 7 marks]


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Question 11 (Mendelian Genetics)

The figure below shows the pedigrees (family trees) of two families, A and B, representing the occurrence of colour blindness in parents and their children.

(a) In individuals A4 and B3 were to produce children, what proportions of different off spring would you expect to find?

Male colour blind

Male, full colour vision

Female colour blind

Female full colour vision

[1]

(b) Redraw pedigree A only showing all possible genotypes of the individuals in each generation. ( mark each individual)

[4]

Another human characteristic showing this pattern of inheritance is haemophilia.

(c) Suggest two reasons why haemophilia is less common than colour blindness. [2]

The table shows the results of a cross between homozygous recessive black bodied, bent winged fruit flies (bb, nn) and the double heterozygote (Bb, Nn).

Phenotype Number Genotype Combination of traits Normal body, normal wing 83 BbNn Parental Black body, normal wing 71 bbNn Non-parental Normal body, bent wing 69 Bbnn Non-parental Black body, bent wing 85 bbnn Parental


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(d) What is the expected phenotype ratio of the offspring? [1] Use the chi-squared test to check whether this assumption is justified.

(e) What is the Null hypothesis for this data? There is no significant difference* between the observed numbers and the expected numbers, any difference is due to chance*.

[1]

(f) What is the expected number in each class? [1]

(g) With all working shown clearly, show your calculations for chi-square below. [2]

Distribution of 2 Degrees of

freedom Probability, p

0.10 0.05 0.02 0.01 0.001 1 2.71 3.84 5.41 6.64 10.83 2 4.61 5.99 7.82 9.21 13.82 3 6.25 7.82 9.84 11.35 16.27 4 7.78 9.49 11.67 13.28 18.47

(h) Explain the conclusion that can be draw from your your calculated 2 value and the table above.

[2]

[Total: 15 marks]

Question 12 (9700 Nov08/P4/Q8) (Mendelian Genetics) In mice there are several alleles of the gene that controls the intensity of pigmentation of the fur. The alleles are listed below in order of dominance with C as the most dominant.

C = full colour Cch = chinchilla Ch = himalayan Cp = platinum Ca = albino

The gene for eye colour has two alleles. The allele for black eyes, B, is dominant, while the allele for red eyes, b, is recessive. A mouse with full colour and black eyes was crossed with a himalayan mouse with black eyes. One of the offspring was albino with red eyes.

Using the symbols above, draw a genetic diagram to show the genotypes and phenotypes of the offspring of this cross. [6]

[Total: 6 marks]

2 (O E )2

E


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Question 13 (Mendelian Genetics)

The parental generation consisting of crosses between white-eyed and normal-eyed forms of fruit fly (Drosophila sp.) and the next two subsequent generations are shown below. A begins with a cross between a normal-eyed female and a white-eyed male while B begins with a reciprocal cross.

(The genotypic ratios obtained in A and B are as follows: first generation offspring 1:1, second generation offspring 1:1:1:1) (a) What is meant by reciprocal cross? [1]

(b) Explain the difference in the first generation male obtained in A and B. [3]

(c) Using appropriate symbols and genetic diagrams, explain the the cross between the first generation offspring seen in A and B. [5]

(d) Explain how would you find out which female in the second generation of A is heterozygous for the eye colour. [2]

[Total: 9 marks]

Parental generation

Cross between first generation offspring respectively

Generation 2


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Question 14 (Proteins and Enzymes)

Consider the following hypothetical protein (Fig. 14.1) that spans the membrane of a cell:

Fig. 14.1

(a) (i) What class of macromolecules does the molecules (except for the protein) that constitute the above membrane belong to? [1]

(ii) Explain the important features of these molecules that allow them to form membranes. [2]

(b) Consider the interaction between the side chains of valine 141 and cysteine 136.

(i) Suggest one possible type of interaction between these two groups. [1]


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(ii) Could changing valine 141 to a different amino acid strengthen the interaction between the amino acids at position 141 and 136 of this protein? [2]

If no, briefly explain why not. If yes, explain how the interaction will be weakened.

(c) Amino acids 100 through 129 of this protein are all alanines, leucines or valines. What interactions occur between the side chains of these amino acids and the molecules that constitute the inside of the membrane? [1]

(d) If amino acids 100 through 129 are replaced with serine, glutamine and asparagines (all three being amino acids that are hydrophilic), the resulting protein has the same general structure. Would you then expect that this new protein be still found embedded in the membrane? Explain your answer. [2]

[Total: 9 marks]

Question 15 (Proteins and Enzymes)

(a) Write down the structural formula of the tripeptide formed by alanine (R = CH3), glycine (R = H) and serine (R = -CH2OH) joined together in that order. Indicate and name the bond formed between these three amino acids.

[2]

(b) Lipase is an enzyme that catalyses the hydrolysis of triglycerides. It is a soluble globular protein. The function of an enzyme depends upon the precise structure of its tertiary structure. Fig. 15.1 represents the structure of an enzyme. The black strips represent the disulfide bonds which help to stabilize its tertiary structure.

Fig. 15.1

(i) Describe the nature of the disulfide bonds that help stabilize the tertiary structure of a protein such as lipase.

[3]

(ii) Name two other types of bonds that help stabilize the tertiary structure. [1] [Total: 7 marks]


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Microsoft Word - 2014 YR5 Hol Homework Questions

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