T opic 1 Assessment Booklet Marks = 123 Time Allowed 152 minutes Q1. The diagram shows part of a DNA molecule. (a) How many nucleotides are shown in the diagram above? ______________________________________________________________ _____ (1) (b) Name the type of bond labelled X in the diagram. ______________________________________________________________ _____ (1) (c) The enzymes DNA helicase and DNA polymerase are involved in DNA replication. Describe the function of each of these enzymes. DNA helicase _______________________________________________________ ______________________________________________________________ Page 1 of 48
Transcript
T opic 1 Assessment Booklet
Marks = 123 Time Allowed 152 minutes
Q1.The diagram shows part of a DNA molecule.
(a) How many nucleotides are shown in the diagram above?
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 used shortening of strips of muscle tissue caused by contraction as evidence that ATP was being hydrolysed.
• She took four slides A, B, C and D, and added strips of muscle tissue of the same length to each slide.
• She then added the same volume of ATP solutions of different concentrations to the four slides 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.
SlideConcentration
of ATP solution added to slide / × 10–6 mol dm–3
Final length of muscle tissue 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.
A student wanted to produce a dilution series of a maltose solution so he could plot a calibration curve. He had a stock solution of maltose of concentration 0.6 mol dm−3 and distilled water. He made a series of dilutions from 0.1 to 0.6 mol dm−3.
(c) Complete the table below by giving all headings, units and the concentration of the maltose solution produced.
Concentration of maltose solution
Volume of 0.6 mol dm−3 maltose solution / cm3
_______________
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/ ____________ ______ / _______
____________ 5 10
(2)
The student performed the Benedict’s test on six maltose solutions ranging from 0.1 mol dm−3 to 0.6 mol dm−3. He placed a sample of each solution in a colorimeter and recorded the light absorbance.
His results are shown in the graph below.
(d) Explain how you would use the graph to determine the maltose concentration with a light absorbance of 0.45 arbitrary units.
(c) Following digestion and absorption of food, the undigested remains are processed to form faeces in the parts of the intestine below the ileum.
The faeces of people with constipation are dry and hard. Constipation can be treated by drinking lactulose. Lactulose is soluble, but is not digested or absorbed in the human intestine.
Use your knowledge of water potential to suggest why lactulose can be used to help people suffering from constipation.
(d) Lactulose can also be used to treat people who have too high a concentration of hydrogen ions (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)
Q5.A student investigated the effect of lipase concentration on the hydrolysis of lipids.
He took a beaker containing a suspension of lipids. He placed a pH probe attached to a data logger into the beaker. After 5 minutes, he added the lipase solution. The data logger recorded the pH. The apparatus used is shown in the diagram below.
(a) The student did not add a buffer to the lipase solution.
(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 strand of mtDNA has been copied.
A piece of mtDNA is 16 500 base pairs long and is replicated at a rate of 50 nucleotides per second.
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)
Q8.Scientists measured the mean amino acid concentration in white wines made from grapes grown organically and white wines made from grapes that were not grown organically.
(a) Which test could the scientists have used to identify that there are amino acids in white wine?
(d) The scientists used a statistical test to determine whether there was a significant difference in the amino acid concentration in the two types of white wine. They obtained a value for P of 0.04.
Name the statistical test the scientists used and give a reason for your answer.
Was the difference significant? Give a reason for your answer.
Name of statistical test ________________________________________________
Reason for choice ____________________________________________________
(b) Scientists investigated changes in the use of artificial fertiliser in India between 1970 and 2005. They also investigated changes in the fertiliser response ratio. This ratio shows how many kg of grain are produced for each kg of fertiliser used.
The graph shows their results in the form the scientists presented them. (A hectare
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is a unit of area commonly used in agriculture)
Use these data to calculate the difference in the mass of grain produced per hectare in 1970 compared with 2005.
Show your working.
Difference ____________________ kg hectare−1
(2)
(c) Use the data in the graph above to evaluate the use of artificial fertilisers on grain-producing crops in India.
Q13.(a) Most human cells contain two copies of each gene. However, there might be up to
15 copies of the gene for amylase (AMY1). Scientists investigated the number of copies of the AMY1 gene in individual people in two populations. One population had a high-starch diet and the other population had a low-starch diet.
Q14.Scientists investigated how the concentration of protein in blood plasma changes in people between the ages of 60 and 95.
The graph shows the scientists’ results. The bars show ±1 standard deviation.
(a) What is the difference between males and females in the fall in mean concentration of protein in blood plasma between 60 and 95 years?
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Answer = ___________ g dm−3
(1)
(b) Use the graph above to calculate the rate of change of the mean concentration of protein in the blood plasma of males between the ages of 60 and 95.
Show your working.
Answer = ___________ g dm−3 year−1
(2)
(c) What can you conclude from the graph above about the effect of ageing on the mean concentration of protein in the blood plasma in males and females?
(d) The scientists measured the absorption of each sample of blood plasma using a colorimeter. They used a calibration curve to find the concentration of protein in samples of blood plasma.
Describe how the scientists could obtain data to produce a calibration curve and how they would use the calibration curve to find the concentration of protein in a sample of blood plasma.
Q15.In mammals, in the early stages of pregnancy, a developing embryo exchanges substances with its mother via cells in the lining of the uterus. At this stage, there is a high concentration of glycogen in cells lining the uterus.
(d) In humans, after the gametes join at fertilisation, every cell of the developing embryo undergoes 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)
Q16.(a) Give the two types of molecule from which a ribosome is made.
(c) 1. DNA helicase – (unwinding DNA and) breaking hydrogen bonds / bonds between chains / bases / strands;
2. DNA polymerase – joins (adjacent) nucleotides OR forms phosphodiester bond / sugar-phosphate backbone;
1. Accept H bonds.1. Accept hydrolyses for breaks2. Reject forms hydrogen bonds (between nucleotides /
bases)2
(d) 1. ATP has ribose and DNA nucleotide has deoxyribose;2. ATP has 3 phosphate (groups) and DNA nucleotide
has 1 phosphate (group);3. ATP – base always adenine and in DNA nucleotide
base can be different / varies;Both parts of each MP needed3. Reject Uracil / U3. Accept C, T or G for different basesAccept annotated diagram for any of the three marks
2 max[6]
Q2.(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
(b) Adenosine diphosphate and (inorganic) phosphate;Accept ADP for adenosine diphosphateAccept Pi / PO4
3– / P in a circle for inorganic phosphateReject adenine diphosphateReject phosphorus / P for phosphate
1
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(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 qualifiedReject concentration / volume of ATP hydrolase
2 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 more muscle contraction / shortening of muscle;
Accept more ATP available for correct/named aspect of muscle contractionIdea of more is required once.Reject energy produced
2
(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 0.61Accept 5 × 10–6
Accept correct answer however expressedMax 2 for incorrect final answer
3[10]
Q3.(a) Glucose (and glucose);
1
(b) (α1,4) Glycosidic;1
(c) 1. Headings correct – mol dm–3 and volume of water / cm3;
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2. Concentration correct. ie 0.2;2
(d) Line of best fit drawn;
Read off value at 0.45.2
[6]
Q4.(a) (a monomer is a smaller / repeating) unit / molecule from which larger molecules /
polymers are made;Reject atoms / elements / ’building blocks’ for units / moleculesIgnore examples
1
(b) Similarity1. Both contain galactose / a glycosidic bond;
Ignore references to hydrolysis and / or condensation
Ignore alpha / beta prefix for glucoseDifference must be stated, not implied
2
(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 hard
2
(d) (-) 84.1(%);;Accept (-) 84.15(%)Allow 1 mark for84OR
OR
2[7]
Q5.(a) Student was measuring change in pH
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ORBuffer would maintain a constant pH.
1 max
(b) 1. Volume of suspension of lipids;
2. Concentration of suspension of lipids;
3. Volume of lipase solution;
4. Temperature;2 max
(c) Boiled lipase solution;1
(d) –0.34 = 2 marks
0.34 = 1 mark2
(e) 1. Fatty acids produced;
2. Curve levels off as all substrate used up.accept the lower pH inactivates / denatures the enzyme
2
(f) 1. Faster fall in pH and levels off at same point;
2. More enzyme = substrate complexes formed;
3. Same amount of fatty acids produced / product3
[11]
Q6.(a) P – glycerol
Q – fatty acid (chains)Accept phonetic spelling
2
(b) Ester (bond);1
(c) 1. (Mix / shake sample) with ethanol, then water;Sequence is important
2. White / milky (emulsion);Ignore cloudyReject precipitate
2[5]
Q7.(a) 1. Phosphate, deoxyribose and base correctly labelled;
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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 labelsAccept any orientation of diagram, eg inverted / mirror imageAccept any pentagon for deoxyribose
2
(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 accurate replication.
2 max
(c) C. 550 seconds;1
[5]
Q8.(a) Biuret;
Ignore any other detailAccept• Copper sulfate and sodium hydroxide• CuSO4 + NaOH• Alkaline copper sulfate• Copper sulphate and sodium hydroxide• Alkaline copper sulphate• Biurette
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• Buiret• Biruet• BieuretReject burette or Beirut
1
(b) Draw around 1
(c) Nitrogen;Ignore N
1
(d) Choice: (Student’s) t-test;Reason for choice: Looking for differences between two means;
Reason: Allow comparing contrasting two means
Explanation: Difference is significant / not due to chance because the P value is 0.04 / is less than 0.05;
Explanation: Assume ‘it’ means differenceExplanation: Reject result / data is significant / not due to chanceExplanation: do not accept P value is less than 0.04
3[6]
Q9.(a) 1. Polar molecule;
2. Acts as a (universal) solvent;
OR
3. (Universal) solvent;
4. (Metabolic) reactions occur faster in solution;
OR
5. Reactive;
6. Takes place in hydrolysis / condensation / named reaction;Polar molecule so acts as (universal) solvent so (metabolic reactions are faster = 3 marks
4
(b) Name of ion;
Correct function within cell;Ions other than sodium in specification are H+, Fe2+ and PO4
3–
but accept any correct ion (other than sodium) plus relevant function = 2.
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Allow ion to be named in words but not as element, e.g, iron ion but not iron.
2
(c) 1. Comparison: both move down concentration gradient;
2. Comparison: both move through (protein) channels in membrane;Accept aquaporins (for water) and ion channels
3. Contrast: ions can move against a concentration gradient by active transport3
[9]
Q10.(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
(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 bonds3. Accept (at 60 °C) Change in shape of the active site /
active site no longer complementary so no enzyme-substrate complexes 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 /
viral antigens / 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’
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2. Accept ‘no carrier proteins for dipeptides’2
[10]
Q11.(a) Two suitable examples;
Examples1. amino acid / protein / polypeptide / peptide;2. nucleic acid / nucleotide / base;3. DNA;4. RNA;5. ATP / ADP;6. NAD / NADP (reduced or not);7. Cyclic AMP / cAMP;8. Chlorophyll;
List rule appliesReject for either point nitrates / nitrites / ammonia / ammonium / urea4. Accept pre-mRNA / mRNA / rRNA / tRNA
2 max
(b) Correct answer in the range 90 to 133.2 scores 2 marks;1 mark for answers where yield calculated correctly for 1970 OR 2005;(1970 in range) 170.8 to 176.4OR(2005 in range) 266.4 to 304.0;
Accept positive or negative values2
(c) 1. Using more but getting less response over time;2. The graph shows correlation but doesn’t prove changes in yield due to
fertiliser / but there could be other factors;3. Becomes less cost effective with time;
Idea of over time is important1. accept fertiliser becomes less effective over time1. Accept use of figures from graph1. Accept the idea of less grain / crop over time2. Ignore whether correlation is positive or negative
2 max[6]
Q12.(a) 1. Cellulose is made up of β-glucose (monomers) and glycogen is made up of α-
glucose (monomers);2. Cellulose molecule has straight chain and glycogen is branched;3. Cellulose molecule has straight chain and glycogen is coiled;4. glycogen has 1,4- and 1,6- glycosidic bonds and cellulose has only 1,4-
glycosidic bonds;Ignore ref. to H bonds / microfibrils
2 max
(b) Any two from:1. Insoluble (in water), so doesn’t affect water potential;
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2. Branched / coiled / (α-)helix, so makes molecule compact;ORBranched / coiled / (α-)helix so can fit many (molecules) in small area;
3. Polymer of (α-)glucose so provides glucose for respiration;4. Branched / more ends for fast breakdown / enzyme action;5. Large (molecule), so can’t cross the cell membrane
Require feature and explanation for 1 mark1. Accept Ψ or WP1. Accept Insoluble so doesn’t affect osmosis1. Do not allow ref to ‘doesn’t affect water leaving cells4. Ignore ‘surface area’4. Accept ‘branched so glucose readily released’
2 max
(c) Iodine/potassium iodide;1
(d) For correct answer of 40 (μm) award 2 marks;Evidence of division by 500: award 1 mark
Allow tolerance of 0.5mm i.e. 20±0.5mm2
(e) 1. Scanning electron (microscope);2. 3D (image);
Accept SE(M)2. Ignore any other correct features
2[9]
Q13.(a) 1. Low starch, fewer copies;
2. Ranges overlap almost completely;ORRanges overlap from 2 − 13 copies;
3. (surprisingly) very few / 2 or 3% have only 2 copies / are diploid;4. the mode / highest percentage for low starch is 4 copies and for high
starch is 6;5. the range / spread is greater with high starch;
4. “most people” is not equivalent to mode3 max
(b) 1. More mRNA / more transcription;2. More translation / enzyme;3. So reaction faster;
The idea of “more” must be stated at least once.2. Accept ‘amylase’ for enzyme3. “More starch digested” is insufficient
3
(c) 1. Mutation(s) produce extra copies of (AMY1) gene;2. Those with more copies / this adaptation/mutation reproduce / survive
better on high starch diet;2. And pass on multiple copies / this adaptation/mutation (to offspring);
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Ignore ref. to single allele/gene3
[9]
Q14.(a) 6 (g dm−3);
1
(b) Correct answer of (−)0.14;1 mark for correct difference in concentration (5) divided by 35 / (69 − 64) ÷ 35
/ 1 ÷ 7Ignore +/- signIgnore additional d.p.Accept 0.31(4) for 1 mark if female data used
2
(c) 1. Protein content decreases with age and decreases more in females;2. Difference (between sexes) only significant at 95 years because SDs do
not overlap;ORDifferences not significant because 2 × SD would overlap;
2
(d) 1. Produce known concentrations of protein;2. Measure absorbance of each concentration
ORMeasure each concentration with colorimeter;
3. Plot a graph of absorbance on y-axis against concentration (on x-axis) and draw curve;
4. Use absorbance of sample to find protein concentration from curve;1. Idea of known concentrations required.Accept % transmission / absorption for absorbance
3 max
(e) 1. (Lower plasma protein concentration suggests) fewer antibodies;Ignore ref. to other proteins.Reject answers which refer to white blood cells as proteins.
(c) 1. Membrane folded so increased / large surface area;ORMembrane has increased / large surface area for (fast) diffusion / facilitated diffusion / active transport / co-transport;
2. Large number of protein channels / carriers (in membrane) for facilitated diffusion;
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 and reference 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 OR 0.0000328 OR 0.000033);
Accept any number of significant figures as long as rounding correct (3.28125 × 10 −5 scores 2 marks)
2[8]
Q16.(a) 1. One of RNA / ribonucleic acid(s) / nucleotide(s)/nucleic acid(s) / rRNA /
Reject DNA, deoxyribonucleic acid, tRNA, transfer RNA, transfer ribonucleic acid, mRNA, messenger RNA, messenger ribonucleic acid.Ignore enzyme(s), base(s).
1
(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.
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Reference to ‘introns present in mRNA’ disqualifies mp1 but allow ECF for mp2.Accept for 1 mark mRNA contains only exons.
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