Topic One New Spec Class: Date: - missestruch · 2019. 12. 19. · 7KHVDOLYDRIPRVWKXPDQVFR QWDLQV....

Topic One New Spec

Name: ________________________

Class: ________________________

Date: ________________________

Time: 198 minutes

Marks: 153 marks

Comments:

Page 1 of 61Woodford County High School

The saliva of most humans contains α-amylase. The gene encoding α-amylase is called AMY1; itis located on chromosome 1.

As a result of mutation, humans might have more than one copy of the AMY1 gene on one, orboth, of their copies of chromosome 1. A team of scientists investigated whether the number ofcopies of the AMY1 gene was associated with the concentration of α-amylase in the saliva of 58human volunteers.

The graph shows their results. Each circle represents one volunteer.

1

(a) What was the range in the number of copies of the AMY1 gene?

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(1)


(b) The scientists found the mean number of copies of gene AMY1 was 4.4 genes per person.

Four values of the standard deviation of this mean are given below.Estimate which of the four values for the standard deviation is most likely for this mean.Indicate your choice by placing a tick in the appropriate box.

Use evidence from the graph to justify your answer.

± 0.002 ± 0.02 ± 0.20 ± 2.00

Justification_________________________________________________________

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(2)

(c) The scientists calculated a correlation coefficient, R, from their data.They found R = 0.50, with P

(d) The number of copies of the AMY1 gene is unlikely to affect people’s ability to digeststarch.

Explain why.

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(3)

(Total 9 marks)

(a) DNA is a polymer of nucleotides. Each nucleotide contains an organic base.

Explain how the organic bases help to stabilise the structure of DNA.

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(2)

2


(b) Triplets of bases in a DNA molecule code for the sequence of amino acids in a polypeptide.The genetic code is frequently written as the three bases on mRNA that arecomplementary to a triplet on DNA. Table 1 shows what different combinations of bases onmRNA code for. The names of amino acids are abbreviated. For example, ‘Ala’ stands foralanine.

Table 1

First base Second base Third base

Guanine (G) Adenine (A) Cytosine (C) Uracil (U)

G

GGG Ala

GGA Gly

GGC Gly

GGU Gly

GAG Glu

GAA Glu

GAC Asp

GAU Asp

GCG Ala

GCA Ala

GCC Ala

GCU Ala

GUG Val

GUA Val

GUC Val

GUU Val

G

A

C

U

A

AGG Arg

AGA Arg

AGC Ser

AGU Ser

AAG Lys

AAA Lys

AAC Asn

AAU Asn

ACG Thr

ACA Thr

ACC Thr

ACU Thr

AUG Met

AUA Iso

AUC Iso

AUU Iso

G

A

C

U

C

CGG Arg

CGA Arg

CGC Arg

CGU Arg

CAG Gln

CAA Gln

CAC Hist

CAU Hist

CCG Pro

CCA Pro

CCC Pro

CCU Pro

CUG Leu

CUA Leu

CUC Leu

CUU Leu

G

A

C

U

U

UGG Trp

UGA stop

UGC Cyst

UGU Cyst

UAG stop

UAA stop

UAC Tyr

UAU Tyr

UCG Ser

UCA Ser

UCC Ser

UCU Ser

UUG Leu

UUA Leu

UUC Phe

UUU Phe

G

A

C

U

Suggest one advantage of showing the genetic code as base sequences on mRNA, ratherthan triplets on DNA.

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(1)

(c) What name is given to a group of three bases on mRNA that codes for an amino acid?

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(1)


(d) Use information from Table 1 to explain why the genetic code is described as degenerate.

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(2)

(e) Suggest the role of the mRNA base triplets UGA, UAG and UAA.

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(2)

(f) Table 2 shows the sequence of mRNA bases forming part of a single gene.

Table 2

Base on DNAtemplate

Base onmRNA

G U G U A C U G G

Encodedamino acid

Complete Table 2 to show the base sequence of the DNA template from which this mRNAwas transcribed and the encoded amino acid sequence.

(2)

(Total 10 marks)


The UK government pays farmers to leave grassy strips around the edges of fields of crops.These grassy strips contain a variety of plant species. Leaving the strips is an attempt toencourage biodiversity of animals.

(a) Give two reasons why the grassy strips increase the biodiversity of animals.

1. _________________________________________________________________

2. _________________________________________________________________

(2)

3

A group of scientists investigated the effect of grassy strips on the biodiversity of soil animals.

• They divided a field into plots measuring 25 m × 5 m, with a 5-metre-wide grassy strip ofland between each plot.

• Each year, they planted wheat in each of the plots.• In the fifth year, they removed samples of soil from each plot where wheat was growing and

from the grassy strips around them.• They sorted each soil sample by hand for 40 minutes to collect the soil animals within the

sample.

(b) The scientists decided to collect animals from the soil samples for 40 minutes.

Suggest how the scientists decided that 40 minutes was an appropriate time.

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(2)


(c) The table below shows how the scientists published their results. They calculated meanvalues and two times the standard deviation (SD) of the mean.

Two standard deviations above and below the mean includes 95.4% of the data.

Group ofanimals

Mean number of animals per m2

(± 2 × SD)Mean number of species per m2

(± 2 × SD)

Soil under wheatcrop

Soil undergrassy strips

Soil under wheatcrop

Soil undergrassy strips

Beetles 41.2 (± 6.4) 80.1 (± 10.1) 10.0 (± 1.6) 17.3 (± 1.0)

Centipedes 18.4 (± 3.6) 13.5 (± 1.0) 1.8 (± 0.3) 2.1 (± 0.2)

Earthworms 244.5 (± 27.1) 281.2 (± 39.4) 3.8 (± 0.3) 5.1 (± 0.2)

Millipedes 38.4 (± 12.2) 36.2 (± 2.9) 3.5 (± 0.3) 3.2 (± 0.2)

Woodlice 0.0 73.9 (± 8.5) 0.0 2.8 (± 0.2)

It would not be possible to calculate an index of diversity from the results in the table.

Explain why.

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(1)


A summary of this research was published in a farming magazine. The journalist concludedthat creating grassy strips around fields had little effect on the diversity of soil animals.

Do you agree with this conclusion?

Use evidence from the table to justify your answer.

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(4)

(Total 9 marks)

(a) Give the two types of molecule from which a ribosome is made.

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(1)

4

(b) Describe the role of a ribosome in the production of a polypeptide. Do not includetranscription in your answer.

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(3)


(c) The table below shows the base sequence of part of a pre-mRNA molecule from aeukaryotic cell.

Complete the table with the base sequence of the DNA strand from which this pre-mRNAwas transcribed.

DNA

A C G C A U U A U pre-mRNA

(1)

(d) In a eukaryotic cell, the base sequence of the mRNA might be different from the sequenceof the pre-mRNA.

Explain why.

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(2)

(Total 7 marks)

(a) Bacteria are often used in industry as a source of enzymes. One reason is becausebacteria divide rapidly, producing a large number of them in a short time.

Describe how bacteria divide.

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(2)

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(b) Washing powders often contain enzymes from bacteria. These enzymes include proteasesthat hydrolyse proteins in clothing stains.

The graph shows the effect of temperature on a protease that could be used in washingpowder.

Explain the shape of the curves at 50 °C and 60 °C.

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(4)


(c) Some proteases are secreted as extracellular enzymes by bacteria.

Suggest one advantage to a bacterium of secreting an extracellular protease in its naturalenvironment.

Explain your answer.

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(2)

(d) Mammals have some cells that produce extracellular proteases. They also have cells withmembrane-bound dipeptidases.

Describe the action of these membrane-bound dipeptidases and explain their importance.

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(2)

(Total 10 marks)


A student investigated the effect of substrate concentration on the initial rate of an enzyme-catalysed reaction.

She added 10 cm3 of an enzyme solution to 10 cm3 of substrate solutions of differentconcentrations. At 30-second intervals, she tested samples of each mixture for the presence ofsubstrate.

• A – in the absence of an inhibitor.• B – with a competitive inhibitor added to the substrate solution.• C – with a non-competitive inhibitor added to the substrate solution.

Her results are shown in the graph below.

(a) Explain the results without inhibitor (curve A) shown in the graph.

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(2)

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(b) The graph shows that the maximum initial rate of reaction (Vmax) when a competitiveinhibitor was present (curve B) is different from that when a non-competitive inhibitor waspresent (curve C).

Explain this difference.

(4)

(c) The Michaelis constant (Km) is the substrate concentration at which the initial rate ofreaction is half its maximum value (Vmax).

How could you use the Michaelis constant to determine the type of inhibition occurring inan enzyme-catalysed reaction?

Use information from the graph to support your answer.

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(1)

(Total 7 marks)

Scientists investigated treatment of a human bladder infection caused by a species of bacterium.This species of bacterium is often resistant to the antibiotics currently used for treatment.

They investigated the use of a new antibiotic to treat the bladder infection. The new antibioticinhibits the bacterial ATP synthase enzyme.

(a) Place a tick (✔) in the appropriate box next to the equation which represents the reactioncatalysed by ATP synthase.

ATP ⟶ ADP + Pi + H2O

ATP + H2O ⟶ ADP + Pi

ADP + Pi ⟶ ATP + H2O

ADP + Pi + H2O ⟶ ATP

(1)

7


(b) The new antibiotic is safe to use in humans because it does not inhibit the ATP synthasefound in human cells.

Suggest why human ATP synthase is not inhibited and bacterial synthase is inhibited.

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(1)


(c) The scientists tested the new antibiotic on mice with the same bladder infection. Theydivided these mice into three groups, C, R and A.

• Group C was the control (untreated).

• Group R was treated with an antibiotic currently used against this bladder infection.

• Group A was treated with the new antibiotic.

They removed samples from the bladder of these mice after treatment and estimated thetotal number of bacteria in the bladder.

Their results are shown in the graph.


The antibiotics were given to the mice at a dose of 25 mg kg−1 per day.

Calculate how much antibiotic would be given to a 30 g mouse each day.

Show your working.

Answer = ____________________ mg

(2)

(d) Calculate the percentage difference in actual numbers of bacteria in group A comparedwith group R. The actual number of bacteria can be calculated from the log10 value byusing the 10x function on a calculator.

Show your working.

Answer = ____________________ %

(2)

(e) The scientists suggested that people newly diagnosed with this bladder infection should betreated with both the current antibiotic and the new antibiotic.

Explain why the scientists made this suggestion.

Use information from the graph in part (c) and your knowledge of evolution of antibioticresistance in bacteria in your answer.

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(3)

(Total 9 marks)


Figure 1 shows all the chromosomes present in one human cell during mitosis. A scientiststained and photographed the chromosomes. In Figure 2, the scientist has arranged the imagesof these chromosomes in homologous pairs.

Figure 1 Figure 2

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(a) Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis. Explainyour answers.

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2. _________________________________________________________________

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(2)


(b) Tick (✓✓✓✓) one box that gives the name of the stage of mitosis shown in Figure 1.

A Anaphase

B Interphase

C Prophase

D Telophase

(1)

(c) When preparing the cells for observation the scientist placed them in a solution that had aslightly higher (less negative) water potential than the cytoplasm. This did not cause thecells to burst but moved the chromosomes further apart in order to reduce the overlappingof the chromosomes when observed with an optical microscope.

Suggest how this procedure moved the chromosomes apart.

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(2)


(d) The dark stain used on the chromosomes binds more to some areas of the chromosomesthan others, giving the chromosomes a striped appearance.

Suggest one way the structure of the chromosome could differ along its length to result inthe stain binding more in some areas.

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(1)

(e) In Figure 2 the chromosomes are arranged in homologous pairs.What is a homologous pair of chromosomes?

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(f) Give two ways in which the arrangement of prokaryotic DNA is different from thearrangement of the human DNA in Figure 1.

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2. _________________________________________________________________

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(2)

(Total 9 marks)


(a) Formation of an enzyme-substrate complex increases the rate of reaction.

Explain how.

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(2)

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(b) A scientist measured the rate of removal of amino acids from a polypeptide with andwithout an enzyme present. With the enzyme present, 578 amino acids were released persecond. Without the enzyme, 3.0 × 10–9 amino acids were released per second.

Calculate by how many times the rate of reaction is greater with the enzyme present.Give your answer in standard form.

Answer = ____________________ times faster

(2)


Another scientist investigated an enzyme that catalyses the following reaction.

ATP → ADP + PiThe scientists set up two experiments, C and L.

Experiment C used• the enzyme• different concentrations of ATP.

Experiment L used• the enzyme• different concentrations of ATP• a sugar called lyxose.

The scientists measured the rate of reaction in each experiment. Their results are shown in thegraph.


(c) Calculate the rate of reaction of the enzyme activity with no lyxose at 2.5 mmol dm–3 of ATPas a percentage of the maximum rate shown with lyxose.

Answer = ____________________ %

(2)

(d) Lyxose binds to the enzyme.

Suggest a reason for the difference in the results shown in the graph with and withoutlyxose.

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(3)

(Total 9 marks)


In mammals, in the early stages of pregnancy, a developing embryo exchanges substances withits mother via cells in the lining of the uterus. At this stage, there is a high concentration ofglycogen in cells lining the uterus.

(a) Describe the structure of glycogen.

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(2)

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(b) During early pregnancy, the glycogen in the cells lining the uterus is an important energysource for the embryo.

Suggest how glycogen acts as a source of energy.

Do not include transport across membranes in your answer.

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(2)


(c) Suggest and explain two ways the cell-surface membranes of the cells lining the uterusmay be adapted to allow rapid transport of nutrients.

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(2)

(d) In humans, after the gametes join at fertilisation, every cell of the developing embryoundergoes mitotic divisions before the embryo attaches to the uterus lining.

• The first cell division takes 24 hours.

• The subsequent divisions each take 8 hours.

After 3 days, the embryo has a total volume of 4.2 × 10−3 mm3.

What is the mean volume of each cell after 3 days? Express your answer in standard form.

Show your working.

Answer = ____________________ mm3

(2)

(Total 8 marks)


(a) Draw the general structure of an amino acid.

(1)

11

Table 1 shows mRNA codons and the amino acids coded for by each codon. It also shows someproperties of the R group of each amino acid.

Table 1

Key to the properties of the R group of each amino acid

No overall change Positively charged Negatively charged


(b) The genetic code is described as degenerate.

What is meant by this? Use an example from Table 1 to illustrate your answer.

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(2)

A scientist investigated changes in the amino acid sequence of a human enzyme resulting frommutations. All these amino acid changes result from single base substitution mutations.This enzyme is a polypeptide 465 amino acids long.

Table 2 shows the result of three of the base substitutions.

Table 2

Amino acidnumber

Correct aminoacid

Amino acid inserted as aresult of mutation

203 Val Ala

279 Glu Lys

300 Glu Lys


(c) What is the minimum number of bases in the gene coding for this polypeptide?

Answer = ____________________

(1)

(d) Use information from Table 1 to tick (✔) one box that shows a single base substitutionmutation in DNA that would result in a change from Val to Ala at amino acid number 203

CAA → CGA

GUU → GCA

GUU → GUC

CAC → CGG

(1)


(e) A change from Glu to Lys at amino acid 300 had no effect on the rate of reaction catalysedby the enzyme. The same change at amino acid 279 significantly reduced the rate ofreaction catalysed by the enzyme.

Use all the information and your knowledge of protein structure to suggest reasons for thedifferences between the effects of these two changes.

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(3)

(Total 8 marks)

(a) Describe the gross structure of the human gas exchange system and how we breathe inand out.

(6)

12

(b) Mucus produced by epithelial cells in the human gas exchange system containstriglycerides and phospholipids.

Compare and contrast the structure and properties of triglycerides and phospholipids.

(5)

(c) Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with thesugar, lactose, attached.

Describe how lactose is formed and where in the cell it would be attached to a polypeptideto form a glycoprotein.

(4)

(Total 15 marks)


(a) Describe the role of two named enzymes in the process of semi-conservative replication ofDNA.

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(3)

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(b) Scientists investigated the function of a eukaryotic cell protein called cyclin A. This proteinis thought to be involved with the binding of one of the enzymes required at the start ofDNA replication.

The scientists treated cultures of cells in the following ways.C – Control cells, untreatedD – Added antibody that binds specifically to cyclin AE – Added RNA that prevents translation of cyclin AF – Added RNA that prevents translation of cyclin A and added cyclin A protein

They then determined the percentage of cells in each culture in which DNA was replicating.


Their results are shown in the table.

Cell treatmentPercentage of cells where DNA

was replicating

CControl

91

DAntibody that binds specifically tocyclin A

11

ERNA that prevents translation ofcyclin A

10

FRNA that prevents translation ofcyclin A and added cyclin A protein

92

Suggest explanations for the results in the table.

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(3)

(Total 6 marks)


(a) What is a monomer?

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(1)

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(b) Lactulose is a disaccharide formed from one molecule of galactose and one molecule offructose.

Other than both being disaccharides, give one similarity and one difference between thestructures of lactulose and lactose.

Similarity ___________________________________________________________

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Difference __________________________________________________________

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(2)


(c) Following digestion and absorption of food, the undigested remains are processed to formfaeces in the parts of the intestine below the ileum.

The faeces of people with constipation are dry and hard. Constipation can be treated bydrinking lactulose. Lactulose is soluble, but is not digested or absorbed in the humanintestine.

Use your knowledge of water potential to suggest why lactulose can be used to help peoplesuffering from constipation.

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(2)

(d) Lactulose can also be used to treat people who have too high a concentration of hydrogenions (H+) in their blood.

The normal range for blood H+ concentration is 3.55 × 10–8 to 4.47 × 10–8 mol dm–3

A patient was found to have a blood H+ concentration of 2.82 × 10–7 mol dm–3

Calculate the minimum percentage decrease required to bring the patient’s blood H +

concentration into the normal range.

Answer = ____________________

(2)

(Total 7 marks)


The diagram represents a triglyceride.15

(a) Name the molecules represented in the diagram by:

Box P _____________________________________________________________

Box Q _____________________________________________________________

(2)

(b) Name the type of bond between P and Q in he diagram.

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(1)

(c) Describe how you would test a liquid sample for the presence of lipid and how you wouldrecognise a positive result.

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(2)

(Total 5 marks)


Read the following passage.

Sizes of populations of normal intestinal bacteria are usually controlled byT cells that are produced slowly and in small numbers by the immune system.These T cells do not normally survive for very long. As a result, they do notrelease large amounts of cytokines. Cytokines are chemicals that can causeswelling of the lining of the intestines. 5

Crohn’s disease is a long-lasting disease that causes swelling of the lining ofthe intestines. It is believed that Crohn’s disease can be caused by a loss oftolerance to normal intestinal bacteria, as shown by an unusually largeresponse by T cells. This response can be triggered by pathogenic bacteria inthe intestines of people with a genetic tendency to Crohn’s disease. 10

Some people’s Crohn’s disease can be controlled by a drug called5-aminosalicylic acid (5-ASA) that reduces swelling. Another drug called6-mercaptopurine (6-MP) may also be used. 6-MP inhibits an enzyme requiredto make adenine and guanine. This is effective because most cells can recyclenucleotides, but T cells are not able to do so. 15

Use information from the passage and your own knowledge to answer the questions.

16


(a) The Crohn’s disease symptom of swelling of the lining of the intestines could be triggeredby pathogenic bacteria in the intestines (lines 6–10).

Suggest how.

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(3)


(b) Suggest the meaning of ‘a genetic tendency to Crohn’s disease’ (line 10).

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(2)

(c) Suggest why 5-ASA is only effective in controlling the swelling of the lining of the intestinesin some people with Crohn’s disease (lines 11–12).

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(2)


(d) Suggest why 6-MP can be used to control the symptoms of Crohn’s disease (lines 13–15).

Do not include details of enzyme inhibition or protein synthesis in your answer.

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(3)

(Total 10 marks)


(a) Draw and label a single DNA nucleotide.

(2)

17

(b) Give two features of DNA and explain how each one is important in the semi-conservativereplication of DNA.

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2. _________________________________________________________________

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(2)


(c) Replication of mitochondrial DNA (mtDNA) is different from that of nuclear DNA.

The replication of the second strand of mtDNA only starts after two-thirds of the first strandof mtDNA has been copied.

A piece of mtDNA is 16 500 base pairs long and is replicated at a rate of 50 nucleotides persecond.

Tick (✓✓✓✓) the box that shows how long it would take to copy this mtDNA.

A 330 seconds

B 440 seconds

C 550 seconds

D 660 seconds

(1)

(Total 5 marks)

(a) State and explain the property of water that can help to buffer changes in temperature.

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(2)

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(b) Water is used to hydrolyse ATP.

Name the two products of ATP hydrolysis.

1. _________________________________________________________________

2. _________________________________________________________________

(1)

Hydrolysis of ATP is catalysed by the enzyme ATP hydrolase.

A student investigated the effect of ATP concentration on the activity of ATP hydrolase. She usedshortening of strips of muscle tissue caused by contraction as evidence that ATP was beinghydrolysed.

• She took four slides A, B, C and D, and added strips of muscle tissue of the same length toeach slide.

• She then added the same volume of ATP solutions of different concentrations to the fourslides and left each slide for five minutes.

• She then recorded the final length of each strip of muscle tissue.

Her results can be seen in the table.

Slide

Concentration ofATP solution addedto slide / × 10–6 mol

dm–3

Final length of muscletissue after 5 minutes / mm

A 2 36

B 4 31

C 6 29

D 8 26

(c) Other than those given, name two variables the student should have controlled.

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2. _________________________________________________________________

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(2)


(d) Describe and explain the pattern shown by the data in the table.

Description __________________________________________________________

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Explanation _________________________________________________________

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(2)

(e) The hydrolysis of 1 dm3 of a 1 mol dm–3 solution of ATP releases 30 500 J of energy.

60% of the energy released during the hydrolysis of 1 mol dm–3 of ATP is released as heat;the rest is used for muscle contraction.

The student added 0.05 cm3 of ATP solution to slide D.

Calculate the energy available from ATP for contraction of the muscle on this slide.

Answer = ____________________ J

(3)

(Total 10 marks)


Mark schemes

(a) 2 to 11;11

(b) ± 2.0;

Data show great variation (around mean)OR4.4 ± 2 × SD includes most of the values measured.

2

(c) 1. Shows a positive correlation;

2. Their probability of getting this correlation by chance is less than 0.0001;

Allow less than 0.01%

3. Correlation is highly significant.

Reject ‘results’ are significant / not due to chance3

(d) 1. Little digestion of starch by salivary amylaseORstarch in mouth for a short periodORsalivary amylase inactivated by stomach acid;

2. Amylase also secreted by pancreas;

3. So (most) starch digestion occurs in small intestine.3

[9]

(a) 1. Hydrogen bonds between the base pairs holds two strands together

2. Many hydrogen bonds provides strength

Reject strong hydrogen bonds2

2

(b) (Because) ribosomes assemble polypeptides using mRNA codeORDNA has two strands each with a different (complementary) base sequence;

1

(c) Codon;1

(d) 1. (Because) some amino acids have more than one codon / mRNA code;

2. Correct example from table.2


(e) 1. Stop translation;

2. Result in detachment of polypeptide chain from ribosome.2

(f)

CAC ATG ACC

Val Tyr Trp

Mark each row2

[10]

(a) Any two valid reasons;

e.g.

1. Increase in plant diversity leads to more types of food for animals;

2. Increase in variety of animals leads to increase in predator species;

3. Increase in niche / habitat2 max

3

(b) 1. Repeat soil sorting for different times and record number of species collected;

2. Find optimum time / time beyond which further sorting does not lead to increase inanimal species found

2

(c) 1. No data on number of individuals in each

species / diversity index

1


(d) Principle:

1. Overlap of 2 × SD shows probability of differences (in means) being due to chance isgreater than 0.95;

Allow converse of MP1

Credit MP1 wherever it appears

Agree:

2. No difference in number of earthworms and millipedes (per m2)

3. No difference in number of species of centipedes or millipedes.

Disagree:

4. More beetles and woodlice in grassy strips;

5. More species of beetles, earthworms, woodlice in grassy strips.4 max

[9]

(a) 1. One of RNA / ribonucleic acid(s) / nucleotide(s)/nucleic acid(s) / rRNA / ribosomalRNA / ribosomal ribonucleic acidandone of protein(s) / polypeptide(s) / amino acid(s) / peptide(s) / ribosomalprotein;

Reject DNA, deoxyribonucleic acid, tRNA, transfer RNA, transferribonucleic acid, mRNA, messenger RNA, messenger ribonucleicacid.

Ignore enzyme(s), base(s).1

4

(b) 1. mRNA binds to ribosome;2. Idea of two codons / binding sites;3. (Allows) tRNA with anticodons to bind / associate;4. (Catalyses) formation of peptide bond between amino acids (held by tRNA

molecules);5. Moves along (mRNA to the next codon) / translocation described;

Assume ‘it’ refers to ribosome.3 max

(c) TGCGTAATA;Any errors = 0 marks

1


(d) 1. Introns (in pre-mRNA);2. Removal of sections of (pre-mRNA) / splicing;

Introns removed’ scores 2 marks.

Reference to ‘introns present in mRNA’ disqualifies mp1 but allowECF for mp2.

Accept for 1 mark mRNA contains only exons.2

[7]

(a) 1. Binary fission;2. Replication of (circular) DNA;3. Division of cytoplasm to produce 2 daughter cells;4. Each with single copy of (circular) DNA;

1. Ignore reference to ‘chromosome’

2. Ignore ‘copy’.

4. Ignore references to number of plasmids.2 max

5

(b) 1. Both denatured (by high temperature);2. Denaturation faster at 60 °C due to more (kinetic) energy;3. Breaks hydrogen / ionic bonds (between amino acids / R groups);4. Change in shape of the active site / active site no longer complementary so

fewer enzyme-substrate complexes formed / substrate does not fit;

3. Ignore references to disulphide bonds

3. Accept (at 60 °C) Change in shape of the active site / activesite no longer complementary so no enzyme-substratecomplexes formed / substrate does not fit;

4

(c) 1. To digest protein;2. (So) they can absorb amino acids for growth / reproduction / protein synthesis /

synthesis of named cell component;OR(So) they can destroy a toxic substance / protein;

1. For ‘digest’ accept ‘break down’ here.

2. Accept ‘(so) they can destroy antibodies / antibiotics / viralantigens / bacterial antigens’

2


(d) 1. Hydrolyse (peptide bonds) to release amino acids;2. Amino acids can cross (cell) membrane;

ORDipeptides cannot cross (cell) membrane;ORMaintain concentration gradient of amino acids for absorption;OREnsure (nearly) maximum yield from protein breakdown;

2. Ignore references to crossing gut membranes.

2. Accept ‘there are carrier proteins for amino acids’

2. Accept ‘no carrier proteins for dipeptides’2

[10]

(a) 1. Increases because more enzyme-substrate complexes formed;

Neutral; more collisions

2. Levels off because all enzyme molecules involved in enzyme-substrate complexes (ata given time)

1. and 2. Accept ES

2. Reject enzymes are used upORLevels off because no free active sites (at a given time)ORLevels off because enzyme (concentration) is limiting factor.

2

6

(b) 1. Competitive inhibitor binds to active sites of enzyme but non-competitive inhibitorbinds at allosteric site / away from active site;

2. (Binding of) competitive inhibitor does not cause change in shape of active site but(binding of) non-competitive does (cause change in size of active site);

3. So with competitive inhibitor, at high substrate concentrations (active) enzyme stillavailable but with non-competitive inhibitor (active) enzymes no longer available;

4. At higher substrate concentrations likelihood of enzyme-substrate collisions increaseswith competitive inhibitor but this is not possible with non-competitive inhibitor;

4

(c) Reaction with non-competitive inhibitor has the same value of Km as with no inhibitor /

value is 5 (g dm–3) / reaction with competitive inhibitor has higher Km value than with no

inhibitor / value is 7 (g dm–3).1

[7]


(a)

ATP ⟶ ADP + Pi + H2O

ATP + H2O ⟶ ADP + Pi

ADP + Pi ⟶ ATP + H2O

ADP + Pi + H2O ⟶ ATP

1

7

(b) 1. Human ATP synthase has a different tertiary structure to bacterial ATP synthaseORHuman ATP synthase has a different shape active site to bacterial ATPsynthaseORAntibiotic cannot enter human cells/mitochondriaORAntibiotic not complementary (to human ATP synthase);

1

(c) 0.75;One mark for showing 30 g = 0.03 kg;One mark for showing 0.025 mg g−1

2

(d) Answer in range 97.0 − 97.8%;ORAnswer in range 3288 − 4368%;

2. 1 mark for correct log10 readings from graph converted toactual numbers

(16.98 − 19.50 and 660.7 − 758.6)2


(e) 1. (From the graph in part c) New / old antibiotic does not kill all bacteria;OR(From the graph in part c) Some bacteria are resistant to the new / oldantibiotic;

2. Resistant bacteria will reproduce to produce (more) resistant bacteria;3. (Use of both) one antibiotic will kill bacteria resistant to the other antibiotic;

ORUnlikely that bacteria are resistant to both the new and the old antibiotic;ORUse of both antibiotics (likely to) kill all / most bacteria;

Accept use of ‘ A’ for ‘new antibiotic’ and ‘ R’ for ‘old antibiotic’.

1. Must relate to the bacteria that are still present – ‘somebacteria are killed’ or ‘the bacteria number is reduced ’ isinsufficient.

2. Accept ‘resistant bacteria reproduce to pass on resistancegene / allele’

3. ‘Use of both antibiotics will be more effective’ is insufficient.3

[9]

(a) 1. The (individual) chromosomes are visible because they have condensed;

Both parts of each answer are required – evidence and explanation.

For ‘they’ accept ‘chromosomes/chromatin/DNA’

Accept ‘tightly coiled’ or ‘short and thick’ for condensed but do notaccept ‘contracted’.

Ignore references to nucleus/nucleolus/nuclear membrane.

2. (Each) chromosome is made up of two chromatids because DNA has replicated;


Accept ‘sister chromatids’ for ‘two chromatids’.

Ignore references to nucleus/nucleolus/nuclear membrane.

3. The chromosomes are not arranged in homologous pairs, which they would be if itwas meiosis;


Accept not meiosis because bivalents/chiasmata/crossing over notseen.

Ignore references to nucleus/nucleolus/nuclear membrane.2 max

8

(b) Automarked q – ✔ prophase1


(c) 1. Water moves into the cells/cytoplasm by osmosis;

Reject water moving into chromosomes/nucleus.

2. Cell/cytoplasm gets bigger;

Accept idea of cell/cytoplasm has greater volume/swells/expands.

Ignore references to pressure changes, turgidity and chromosomesbeing more dilute.

Ignore references to changing water/fluid contents of the cell.

Allow ECF for ‘nucleus expands’ but not for ‘chromosomes expand’.2

(d) Differences in base sequences

OR

Differences in histones/interaction with histones

OR

Differences in condensation/(super)coiling;

Answer must be in context of differences in arrangement ofchromosomes not just related to the properties of the stain.

Accept spec section 8 ideas e.g. different methylation/acetylation

Accept different genes

Reject different alleles1

(e) (Two chromosomes that) carry the same genes;

Reject ‘same alleles’

Accept ‘same loci’ (plural) or ‘gene s for the same characteristics’1

(f) (Prokaryotic DNA) is

1. Circular (as opposed to linear);

2. Not associated with proteins/histones ;

3. Only one molecule/piece of DNAORpresent as plasmids;

Max 1 if prokaryotic DNA only found as plasmids OR if prokaryoticDNA is single stranded.

Ignore references to nucleus, exons, introns or length of DNA. Donot credit converse statements.

Ignore descriptions of eukaryotic DNA alone.2 max

[9]


(a) 1. Reduces activation energy;

Accept ‘reduces E a’.

2. Due to bending bondsORWithout enzyme, very few substrates have sufficient energy for reaction;

Accept ‘Due to stress/pressure/tension on bonds’ OR ‘Due toweakening bonds’.

Ignore references to ‘breaking bonds’.2

9

(b) 1.93 × 1011;;

Allow 1 max for

578/3.0 × 10–9

1.93 × 10x when x ≠11

Correct answer with incorrect standard form e.g. 19.3 × 1010

Accept any number of significant figures ≥2, if rounding correct(1.926• × 1011). Same principle applies to one max answers.

2

(c) 31.4;;

Allow 1 max for

0.44 and 1.4

32.8

33.1

30

29.3

Accept any number of significant figures ≥2, if rounding correct(31.4284714).

Same principle applies to 1 max answers.

32.8 = Both readings at 2.5 mmol dm–3 (0.44/1.34)

33.1 = Both readings at 2.5 mmol dm–3 (0.44/1.33)

30 = Incorrect reading for C (0.42/1.4)

29.3 = Incorrect reading for C (0.41/1.4)2


(d) 1. (Binding) alters the tertiary structure of the enzyme ;

Max 1

if lyxose acting as an inhibitor

OR if answer linked to lower rate of reaction

OR if lyxose used an energy source/respiratory substrate

2. (This causes) active site to change (shape);

3. (So) More (successful) E-S complexes form (per minute)

OR

E-S complexes form more quickly

OR

Further lowers activation energy;

Accept ‘acts as a co-enzyme’

Accept description for E-S complexes.3

[9]

(a) 1. Polysaccharide of α-glucose;ORpolymer of α-glucose;

2. (Joined by) glycosidic bondsORBranched structure;

2

10

(b) 1. Hydrolysed (to glucose);2. Glucose used in respiration;

1. Ignore ‘Broken down’

2. ‘Energy produced’ disqualifies mp22

(c) 1. Membrane folded so increased / large surface area;ORMembrane has increased / large surface area for (fast) diffusion / facilitateddiffusion / active transport / co-transport;

2. Large number of protein channels / carriers (in membrane) for facilitateddiffusion;

3. Large number of protein carriers (in membrane) for active transport;4. Large number of protein (channels / carriers in membrane) for co-transport;

1. Accept ‘microvilli to increase surface area’

1. Reject reference to villi.

Note feature and function required for each marking point andreference to large / many / more.

List rule applies.2 max


(d) 3.3 × 10−5 OR 3.28 × 10−5 OR 3.281 × 10−5;1 mark forEvidence of 128 (cells)Correct numerical calculation but not in standard form gains 1 mark (0.00003281 OR0.0000328 OR 0.000033);

Accept any number of significant figures as long as rounding correct(3.28125 × 10 −5 scores 2 marks)

2

[8]

(a)

Accept other correct representations.1

11

(b) 1. More than one codon codes for a single amino acid;

Accept ‘triplet’ or ‘sequence of 3 bases/nucleotides’ for ‘codon’.

Reject ‘production/produces’ for ‘codes for’.

Do not infer mp1 from mp2.

2. Suitable example selected from Table 1;2

(c) 1395;

Accept 1398 and 1401 (for those that include start and/or stopcodons)

Allow 2796 or 2802 or 2790

Ignore ‘bases/base pairs/bp/bps’ written after the numerical answer.1

(d) ✔CAA → CGA1


(e) 1. (Both) negatively charged to positively charged change in amino acid;

2. Change at amino acid 300 does not change the shape of the active siteORChange at amino acid 300 does not change the tertiary structure OR Change atamino acid 300 results in a similar tertiary structure;

Reference to ‘shape’ of active site only needed once.

3. Amino acid 279 may have been involved in a (ionic, disulfide or hydrogen) bond andso the shape of the active site changesORAmino acid 279 may have been involved in a (ionic, disulfide or hydrogen) bond andso the tertiary structure changed;ORAmino acid 279 may be in the active site and be required for binding the substrate;

Reference to ‘shape’ of active site only needed once.

Both parts are required for each mark option.

For ‘a bond’ reject peptide bond.3

[8]

(a) 1. Named structures – trachea, bronchi, bronchioles, alveoli;

Reject mp1 if structures from other physiological systems arenamed but award mp2 if the correct structures are in the correctorder.

2. Above structures named in correct orderORAbove structures labelled in correct positions on a diagram;

Reject mp1 if structures from other physiological systems arenamed but award mp2 if the correct structures are in the correctorder.

3. Breathing in – diaphragm contracts and external intercostal muscles contract;

4. (Causes) volume increase and pressure decrease in thoracic cavity (to belowatmospheric, resulting in air moving in);

For thoracic cavity accept ‘lungs’ or ‘thorax’.

Reference to ‘thoracic cavity’ only required once.

5. Breathing out - Diaphragm relaxes and internal intercostal muscles contract;

Accept diaphragm relaxes and (external) intercostal muscles relaxand lung tissue elastic (so recoils).

6. (Causes) volume decrease and pressure increase in thoracic cavity (to aboveatmospheric, resulting in air moving out);

For thoracic cavity accept ‘lungs’ or ‘thorax’.

Reference to ‘thoracic cavity’ only required once.

If idea of thoracic cavity is missing or incorrect, allow ECF for markpoint 6.

6

12


(b) 1. Both contain ester bonds (between glycerol and fatty acid);

All statements must be clearly comparative or linked by thecandidate, not inferred from separate statements.

Accept mark points shown on adjacent annotated diagrams.

2. Both contain glycerol;

3. Fatty acids on both may be saturated or unsaturated;

4. Both are insoluble in water;

5. Both contain C, H and O but phospholipids also contain P;

Must relate to element.

6. Triglyceride has three fatty acids and phospholipid has two fatty acids plus phosphategroup;

7. Triglycerides are hydrophobic/non-polar and phospholipids have hydrophilic andhydrophobic region;

Accept ‘non-polar’ for hydrophobic and ‘polar’ for hydrophilic.

8. Phospholipids form monolayer (on surface)/micelle/bilayer (in water) but triglyceridesdon’t;

5 max

(c) 1. Glucose and galactose;

Ignore α or β for glucose

2. Joined by condensation (reaction);

3. Joined by glycosidic bond;

4. Added to polypeptide in Golgi (apparatus);;4

[15]

(a) 1. (DNA) helicase causes breaking of hydrogen/H bonds (between DNA strands);

Reject ‘helicase hydrolyses hydrogen bonds’.

2. DNA polymerase joins the (DNA) nucleotides;

Reject if suggestion that DNA polymerase joins the complementarynucleotides or forms H bonds.

Reject if joining RNA nucleotides or forming RNA.

3. Forming phosphodiester bonds;3

13


(b) 1. (Treatment D Antibody binds to cyclin A so) it cannot bind to DNA/enzyme/initiateDNA replication;

For ‘bind to enzyme’ accept ‘activate’.

Idea of ‘initiate DNA replication’ must be linked to start not just lessreplication.

For ‘enzyme’ accept named enzyme.

2. (Treatment E) RNA interferes with mRNA/tRNA/ribosome/polypeptide formation (socyclin A not made);

3. In Treatment F added cyclin A can bind to DNA/enzyme (to initiate DNA replication)ORTreatment F shows that it is the cyclin A that is being affected in the other treatmentsORTreatment F shows that cyclin A allows the enzyme to bind (to DNA)OR(Some cells in D or E) can continue with DNA replication because they have adifferent cyclin A alleleOR(Some cells in D or E) can continue with DNA replication because the antibody/RNAhas not bound to all the cyclin A protein/mRNAOR(Some cells in E) can continue with DNA replication because they contain previouslytranslated cyclin A;

Context needed for Treatment F but it does not need to be named.

For ‘enzyme’ accept named enzyme.3

[6]

(a) (a monomer is a smaller / repeating) unit / molecule from which larger molecules / polymersare made;

Reject atoms / elements / ’building blocks’ for units / molecules

Ignore examples1

14

(b) Similarity1. Both contain galactose / a glycosidic bond;

Ignore references to hydrolysis and / or condensation

Difference2. Lactulose contains fructose, whereas lactose contains glucose;

Ignore alpha / beta prefix for glucose

Difference must be stated, not implied2

(c) 1. (Lactulose) lowers the water potential of faeces / intestine / contents of the intestine;

Accept Ψ for water potential

2. Water retained / enters (due to osmosis) and softens the faeces;

Accept descriptions of soft faeces, eg faeces is less dry / less hard2


(d) (-) 84.1(%);;

Accept (-) 84.15(%)

Allow 1 mark for

84

OR

OR

2

[7]

(a) P – glycerolQ – fatty acid (chains)

Accept phonetic spelling2

15

(b) Ester (bond);1

(c) 1. (Mix / shake sample) with ethanol, then water;

Sequence is important

2. White / milky (emulsion);

Ignore cloudy

Reject precipitate2

[5]

(a) 1. (Presence of) antigen of the (pathogenic) bacteria;

Assume bacteria are pathogenic unless otherwise stated

2. (Causes) more T cells produced / faster T cell production;

3. Against (the pathogen and) normal bacteria;

4. (Long lasting as) cells do not die / live for longer;

5. (More) cytokines / chemicals causing swelling are produced;3 max

16

(b) 1. (Some people) have a mutation / allele / gene;

2. (That) increases the chances / risk / makes it more likely for / causes them to have anunusually large T cell response;OR(That) lowers / removes tolerance to (normal) intestinal bacteria;

2


(c) 1. (Some people might) produce (very) large amounts of cytokine / have large amountsof swelling;

2. (That) 5-ASA drugs cannot control / reduce;

OR

3. Some people may be allergic to / cannot tolerate 5-ASA;

4. So cannot take it;

Award 1 and 2

OR

Award 3 and 42

(d) 1. (Lack of adenine and guanine) will slow / stop DNA synthesis / replication (in T cells);

2. Affects T cells more as they cannot recycle nucleotides;

Needs idea of more / greater effect.

Accept converse idea that ‘other’ cells not as affected as they canrecycle nucleotides.

3. (6-MP therefore) suppresses / slows the (unusually large) T cell / immune responseOR(6-MP causes) fewer / no T cells (to be) produced;

Accept (6-MP) acts as an immunosuppressant drug

4. (So) less cytokine is produced (and therefore less swelling);3 max

[10]


(a) 1. Phosphate, deoxyribose and base correctly labelled;

Accept P in a circle / Pi / PO43– for phosphate.

Do not accept phosphorus for phosphate.

Do not accept only pentose for deoxyribose.

Ignore references to sugar.

Accept a named base, (eg adenine, thymine, guanine, cytosine).

Do not accept uracil or only letters (eg A, T, G or C).

Ignore labelled bonds

2. Correct shapes and bonds in the correct positions (as shown below);

Accept correct shapes with incorrect labels

Accept any orientation of diagram, eg inverted / mirror image

Accept any pentagon for deoxyribose2

17

(b) 1. Weak / easily broken hydrogen bonds between bases allow two strands to separate /unzip;

may appear in the same feature

2. Two strands, so both can act as templates;

may appear in the same feature

3. Complementary base pairing allows accurate replication;

Allow description of complementary base pairing and accuratereplication.

2 max

(c) C. 550 seconds;1

[5]

(a) 1. (water has a relatively) high (specific) heat capacity;

Ignore numbers relating to heat capacity

2. Can gain / lose a lot of heat / energy without changing temperature;ORTakes a lot of heat / energy to change temperature;

Accept due to H bonding between water molecules2

18


(b) Adenosine diphosphate and (inorganic) phosphate;

Accept ADP for adenosine diphosphate

Accept Pi / PO43– / P in a circle for inorganic phosphate

Reject adenine diphosphate

Reject phosphorus / P for phosphate1

(c) 1. Species / organism the muscle tissue came from;ORThickness / type / source of the muscle tissue;

Ignore surface area of muscle tissue

2. Temperature of the muscle tissue / ATP solution / slides;

Need to be qualified

3. pH of the ATP solution;

Need to be qualified

Reject concentration / volume of ATP hydrolase2 max

(d) Description1. As concentration of ATP increases, length of muscle decreases;

Accept negative correlation

Explanation2. More ATP (hydrolysed by ATP hydrolase), so more energy released, so moremuscle contraction / shortening of muscle;

Accept more ATP available for correct/named aspect of musclecontraction

Idea of more is required once.

Reject energy produced2

(e) 4.88 × 10–6 ;;;

If answer incorrect

EITHER

Allow 1 mark for 0.244

Allow 1 mark for 1.22 × 10–5


OR

Allow 1mark for 12200 / 1.525

Allow 1 mark for 0.61

Accept 5 × 10–6

Accept correct answer however expressed

Max 2 for incorrect final answer3

[10]


Topic One New Spec Class: Date: - missestruch · 2019. 12. 19. · 7KHVDOLYDRIPRVWKXPDQVFR QWDLQV....

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Transcript of Topic One New Spec Class: Date: - missestruch · 2019. 12. 19. · 7KHVDOLYDRIPRVWKXPDQVFR QWDLQV....