2003 Jan Module 1

1.

The table below refers to the structure of different types of nucleic acids. If the feature is present, place a ( ) in the appropriate box and if the feature is absent place a cross ( ) in the appropriate box. Feature

DNA

mRNA

Cytosine present Uracil present Pentose sugar present Is single stranded (Total 4 marks)

2.

Read through the following passage about protein structure, then write on the dotted lines the most appropriate word or words to complete the passage. The tertiary structure of a protein depends on its primary and secondary structure. The primary structure is the ………………………………….. of amino acids, which are joined together by ………………………………….. bonds to form a chain. This type of bond is formed when a ………………………………….. reaction takes place between two amino acids. The chain of amino acids may be folded into an alpha helix, held in shape by ………………………………….. bonds. A number of different types of bonds hold the tertiary and quaternary structure in shape. For example, the two peptide chains in insulin are held together by ………………………………….. bonds which form between the ………………………………….. of certain amino acids. (Total 6 marks)

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3.

Yeast cells grown in culture will divide asexually to form a clone of yeast cells. The diagram below shows this process occurring. C h ro m o so m e s

P a re n t c e ll (a)

D a u g h te r c e lls

Explain what is meant by the term clone. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (2)

(b)

On the diagram of one of the daughter cells, draw in the chromosomes that would be found in the nucleus.

(c)

Suggest two advantages of asexual reproduction to an organism. 1 ..............................................................................................................................… .................................................................................................................................... 2 ..............................................................................................................................… .................................................................................................................................... (2) (Total 5 marks)

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4.

The diagram below represents some of the stages involved in the commercial production of the enzyme protease for use in the manufacture of a biological detergent. B a c te ria l c e lls g ro w n in c u ltu r e

P ro te a s e p ro d u c e d b y b a c te ria a n d s e c re te d in to c u ltu re flu id

B a c te r ia l c e lls f ilte r e d o ff a n d d is c a r d e d

P ro te a se p u rifie d

P ro te a s e in c o rp o ra te d in to g ra n u le s (a)

Explain why proteases are incorporated into biological detergents. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (3)

(b)

The bacteria used in this process normally live in hot water springs where the temperature stays above 45 °C. Suggest two reasons why it is an advantage to use the enzymes from these bacteria in the detergents. 1 …………………………………………….………………………………………. ……………………………………………….……………………………………… 2 …………………………………………….………………………………………. ……………………………………………….……………………………………… (2)

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(c)

When the granules containing the protease come into contact with water they swell and release their contents. Suggest why the enzymes are incorporated into granules during the manufacture of biological detergents. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (2) (Total 7 marks)

5.

The diagram below summarises the steps involved in the semi-conservative replication of DNA. D o u b le stra n d o f D N A S te p 1

E nzym e A

T w o stra n d s s e p a ra te d S te p 2 C o m p le m e n ta ry n u c le o tid e s lin e u p a g a in s t e a c h s tra n d S te p 3

E nzym e B

N u c le o tid e s jo in to fo rm tw o n e w p o ly n u c le o tid e c h a in s S te p 4 T w o id e n tic a l D N A m o le c u le s fo r m e d (a)

Describe how Enzyme A separates the two DNA strands in Step 1. ……………………………………………….……………………………………… ……………………………………………….……………………………………… (1)

(b)

In Step 3 the individual nueleotides are joined up to form a polynucleotide chain by Enzyme B. Name the type of reaction that Enzyme B catalyses. ……………………………………………….……………………………………… (1)

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(c)

Give the phase of the cell cycle during which DNA replication occurs. ……………………………………………….……………………………………… (1)

(d)

Draw and label a diagram to show the appearance of a chromosome as it appears in metaphase of mitosis.

(3) (Total 6 marks)

6.

The diagram below shows the structure of the cell surface membrane.

C a rb o h y d ra te

P r o te in s

G ly c o p ro te in

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(a)

The cell surface membrane is composed of a phospholipid bilayer. Explain why tile phospholipids in the bilayer are arranged with the fatty acid tails pointing inwards and the phosphate heads outwards. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (3)

(b)

The diagram has been magnified three million (3 × 106) times. Calculate the width of the cell surface membrane in tin (micrometres). Show your working.

Answer ………………………… µ m (c)

(3)

State one function of each of the following components of the cell surface membrane. Carbohydrate ……………………………………………………………………….. Protein ……………………………………………………………………………… (2) (Total 8 marks)

7.

Some bacteria were grown in a culture with radioactive amino acids. They used the labelled amino acids to synthesise proteins which were incorporated into their cells. The bacteria were then washed thoroughly and mixed with some white blood cells. The amount of radioactivity taken up by the white blood cells was measured at intervals of two hours for 24 hours.

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The white blood cells were phagocytic and engulfed (took up) the bacteria by a process called phagocytosis. Phagocytosis is a form of endocytosis. This process is illustrated in the diagram below. B a c te riu m

M e m b ra n e e x te n s io n s

P hago som e (p h a g o c y tic v a c u o le )

N u c le u s

W h ite b lo o d c e ll Lysosomes then fuse with the phagosome and release their contents into it. The table below shows the level of radioactivity found inside and outside the white blood cells during the 24 hour period. Time after mixing cells together/hours 0 2 4 6 8 10 12 14 16 18 20 22 24 (a)

(i)

Radioactivity inside white blood cells/arbitrary units 0 16 48 61 70 72 72 71 43 21 10 8 5

Radioactivity outside white blood cells/arbitrary units 80 64 32 19 10 8 8 9 37 59 70 72 75

Describe the structure of a lysosome. ………………………………….…….……………………………………… ………………………………….…….……………………………………… ………………………………….…….……………………………………… ………………………………….…….……………………………………… (2)

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(ii)

Describe the roles of lysosomes. ………………………………….…….……………………………………… ………………………………….…….……………………………………… ………………………………….…….……………………………………… ………………………………….…….……………………………………… (2)

(b)

Describe the changes in the level of radioactivity found inside the white blood cells during the period of 24 hours. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (3)

(c)

Suggest what is happening to the bacteria inside the phagosomes between 10 and 14 hours. ……………………………………………….……………………………………… ……………………………………………….………………………………………

(d)

Explain why the amount of radioactivity increases outside the white blood cells after 14 hours. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (2)

(e)

Suggest why the white blood cells did not take up all the radioactivity. ……………………………………………….……………………………………… ……………………………………………….……………………………………… (1) (Total 11 marks)

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8.

The rate of an enzyme-catalysed reaction can be altered by the presence of an inhibitor. An investigation was carried out into the effect of an inhibitor on enzyme activity in barley root tips. Enzyme activity was measured by finding the rate at which oxygen was used by the root tips. Several groups of students carried out experiments in which the volume of oxygen used by root tips of barley seedlings was measured over a period of 2 hours. Each group of students used 50 root tips from seedlings of the same age. These were placed in the same volume of a pH 6.5 buffer solution and kept at 30 °C. After 2 hours, the same volume of a 1% solution of an inhibitor was added to each set of root tips. The students continued to record the volume of oxygen used for a further hour. The results are shown in the graph below. 6

In h ib ito r a d d e d

5 T o ta l v o lu m e o f o x y g en u sed 4 ( a r b itr a r y u n its ) 3 2 1 0

(a)

0

0 .5

1 .0

1 .5

2 .0

2 .5 T im e /h o u rs

3 .0

Compare the enzyme activity during the first two hours of the experiment with the activity after the inhibitor was added. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (2)

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(b)

The inhibitor used in this experiment was a non-active site-directed inhibitor. Explain what is meant by non-active site-directed inhibition. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (3)

(c)

Suggest what effect increasing the concentration of the inhibitor might have on the rate at which oxygen was used by the root tips. Give an explanation for your answer. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (2)

(d)

Suggest why readings were taken for 2 hours before the inhibitor was added ……………………………………………….……………………………………… ……………………………………………….……………………………………… (1)

(e)

Suggest why the root tips were kept at a temperature of 30 °C throughout this investigation. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (2)

(f)

Explain why a buffer solution was used in this experiment. ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… ……………………………………………….……………………………………… (2) 10

(g)

Suggest one reason why the results from the different groups of students might vary. ……………………………………………….……………………………………… ……………………………………………….……………………………………… (1) (Total 13 marks)

9.

Give an account of the structure and function of the polysaccharides cellulose, starch and glycogen. (Allow three lined pages) (Total 10 marks)

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