2018 paper 2 Flashcards

1
Q

Heat stress decreases the light-dependent reaction of photosynthesis. Explain why this leads to a decrease in the light-independent reaction.

A
  1. (Less/no) ATP;

2. (Less/no) reduced NADP;

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2
Q

Another effect of heat stress is a decrease in the activity of the enzyme rubisco. A decrease in the activity of an enzyme means that the rate of the reaction it catalyses becomes slower.
A decrease in the activity of the enzyme rubisco would limit the rate of photosynthesis.
Explain why.

A
  1. (Less/no) carbon dioxide (reacts) with RuBP; 2. (Less/no) GP;
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3
Q

Where precisely is rubisco found in a cell?

A
  1. Stroma (of/in chloroplast);
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4
Q

The scientists concluded that heat stress reduces the activity of rubisco in plant leaves by affecting rubisco activase.
Use all the information to evaluate their conclusion.

A

. Rubisco activity increases with temperature
OR
Rubisco optimum temperature is above (rubisco activase);
2. (Rubisco) activase activity decreases at high temperatures (allow any temperature
above 25 oC.)
OR
(Rubisco) activase optimum (allow in range)
25 to 30 oC.;
3. (Results/graphs suggest) activase cannot/does not affect activity of rubisco;
4. (Results are) only for cotton;
5. (Results are) for isolated enzymes;
6. No stats test;

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5
Q

here are different types of gene mutation.
Put a tick () in the box next to the statement which describes incorrectly the effect
0 2 . 2

A

Box 2.

An inversion will result in a change in the number of DNA bases.

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6
Q

Describe how alterations to tumour suppressor genes can lead to the development of
of the mutation in an exon of a gene.

A

02.2
1.(Increased) methylation (of tumour suppressor genes);
2.Mutation (in tumour suppressor genes);
3. Tumour suppressor genes are not transcribed/expressed
OR
Amino acid sequence/primary structure altered; 4. (Results in) rapid/uncontrollable cell division;

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7
Q

A type of malignant tumour cell divides every 8 hours.
Starting with one of these cells, how many tumour cells will be present after 4 weeks? Assume none of these cells will die.
Give your answer in standard form.

A
  1. Correct answer of 1.9/1.93 x 1025 = 2 marks;;
  2. Incorrect answer but shows 84 = 1 mark OR
    28 x 3 = 1 mark
    OR
    Incorrect answer but shows 672 divided by 8 = 1 mark;
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8
Q

Suggest one reason why it was important that the student left the apparatus for one
hour after the yeast culture reached a constant temperature.

A
  1. (So the) oxygen is used/absorbed/respired;
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9
Q

During her investigation, the coloured liquid moved to the right. Explain why it moved to the right.

A

. Anaerobic respiration produces carbon dioxide; 2. Increase in pressure/volume (of gas);

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10
Q

The student found that the coloured liquid moved 1.5 cm in 24 hours. The diameter of the lumen (hole) of the capillary tubing was 1 mm.
The volume of a capillary tubing is given by πr2𝑙, where π is 3.14 and 𝑙 = length.
Calculate the volume of gas produced in cm3 hour−1. Show your working.

A
  1. Correct answer in range of

4. 9 x 10-4 to 4.91 × 104 = 2 marks;;

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11
Q

Explain why a log scale is used to record the number of cells

A

Large range/difference/increase in numbers;

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12
Q

Many yeast cells die during the death phase. Suggest one reason why

A
  1. Decrease/no glucose/substrate
    OR
    Increase in ethanol/carbon dioxide/acidity;
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13
Q

Assuming no yeast cells died, calculate the predicted size of the population after
10 hours. Show your working.

A
  1. Correct answer of 298000 or 297766 or 297765.59 or 296826 = 2 marks;;
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14
Q

Describe how you would determine the mean percentage cover for beach grass on a
sand dune.

A
  1. Method of randomly determining position (of quadrats) e.g. random numbers table/generator;
  2. Large number/sample of quadrats;
  3. Divide total percentage by number of quadrats/samples/readings;
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15
Q

The scientists concluded that the results shown in Figure 5 were due to succession taking place.
Use Figure 5 to explain why the scientists reached this conclusion.

A

. Beach grass is the pioneer (species);
2. Pioneers/named species change the (abiotic) environment/habitat/conditions/factors;
3. (So) less hostile for named species
OR
(So) more suitable for named species;
4. Conifer/hardwood trees represent climax community;

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16
Q

Use Figure 5 to explain the results in Figure 6

A
  1. Trees block/reduce (sun)light; 1 Reject ‘blocks’ all of
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17
Q

Use Figure 5 to explain the results in Figure 6.

Using evidence from Figure 6, what can you conclude about the net primary productivity

A
  1. (NPP) remains constant;
  2. GPP/photosynthesis and respiration constant; OR
  3. (NPP) low/decreases;
  4. Less light so less photosynthesis/GPP
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18
Q

All the volunteers were given the same food for 3 days.

Suggest and explain one reason why they were given the same food.

A
  1. Affects water potential (of blood/body);

2. Affects volume of urine (produced/removed);

19
Q

Using Table 1, what can you conclude about the effectiveness of furosemide and CVT
in the removal of excess fluid from the body?

A
  1. Furosemide and CVT more effective than placebo/control/C
    OR
    Furosemide more effective (than CVT);
  2. Correct reference to a significant increase/difference as SD’s do not overlap;
20
Q

Furosemide is sometimes used to treat high blood pressure. Suggest how furosemide would cause a decrease in blood pressure

A
  1. Lower volume of blood;
21
Q

Furosemide inhibits the absorption of sodium and chloride ions from the filtrate produced in the nephrons.
Explain how furosemide causes an increase in the volume of urine produced.

A
  1. Water potential of filtrate/tubule decreased;
  2. Less water (reabsorbed) by osmosis (from filtrate/tubule);
  3. Collecting duct (is where osmosis occurs);
22
Q

How much greater is the flow of blood into the kidneys with furosemide than with 3

A
  1. Accept answers in the range 33840 to 34680; 1
23
Q

in genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios.
Suggest two reasons why.

A
  1. Small sample size;
  2. Fusion/fertilisation of gametes is random; 3. Linked Genes;
  3. Epistasis;
  4. Lethal genotypes;
24
Q

hat would be the genotype of the offspring with dwarf plants and mottled leaves?

A
  1. ttmm;
25
Q

se the information provided to explain the results in Table 3.

A
  1. Genes are linked;
  2. Produces few(er) tall, mottled and dwarf, normal offspring;
  3. Crossing over (has occurred);
26
Q

Complete Table 4 to show the expected ratio of phenotypes if the same cross had been carried out but the genes for height of plant and for the type of leaf were on
different homologous pairs of chromosomes.
[2 marks]

A
phenotype of offspring
Tall (plant and) normal (leaves)
Tall (plant and) mottled (leaves)
Dwarf (plant and) normal (leaves)
Dwarf (plant and) mottled (leaves)
Ratio of offspring
9 3 3 1
27
Q

Dopamine is a neurotransmitter released in some synapses in the brain. The transmission of dopamine is similar to that of acetylcholine.
Dopamine stimulates the production of nerve impulses in postsynaptic neurones. Describe how.
Do not include in your answer the events leading to the release of dopamine and the
events following production of nerve impulses at postsynaptic neurones.

A

. (Dopamine) diffuses across (synapse);

  1. Attaches to receptors on postsynaptic membrane;
  2. Stimulates entry of sodium ions and depolarisation/action potential;
28
Q

Dopamine has a role in numerous processes in the brain including pain relief. The release of dopamine can be stimulated by chemicals called endorphins produced in the brain. Endorphins attach to opioid receptors on presynaptic neurones that release dopamine.
Morphine is a drug that has a similar structure to endorphins and can provide pain relief.
Explain how.

A

. Morphine attaches to opioid receptors;

2. (More) dopamine released (to provide pain relief);

29
Q

GABA is a neurotransmitter released in some inhibitory synapses in the brain. GABA causes negatively charged chloride ions to enter postsynaptic neurones.
Explain how this inhibits postsynaptic neurones.

A
  1. (Inside of postsynaptic) neurone becomes more negative/hyperpolarisation/inhibitory postsynaptic potential;
  2. More sodium ions required (to reach threshold)
    OR
    Not enough sodium ions enter (to reach threshold);
  3. For depolarisation/action potential;
30
Q

What is meant by a genome?

A
  1. (All) the DNA in a cell/organism;
31
Q

Explain why the antibody binds to the transcription factor.

A

. (Transcriptional factor/antibody) has a specific/tertiary structure/shape;
2. Complementary (shape/structure);

32
Q

Use Figure 8 to explain what ‘precipitated DNA’ consists of.

A
  1. DNA, transcription factor and antibody;
33
Q

Suggest how single-stranded cDNA could prevent transcription of the P34 gene.

A

Binds to transcription factor gene/DNA

34
Q

Describe the roles of two named types of enzymes used to insert DNA fragments into
plasmids.

A
  1. Restriction (endonuclease/enzyme) to cut plasmid/vector;

2. Ligase joins gene/DNA to plasmid/vector;

35
Q

Suggest two features of the structure of different proteins that enable them to be
separated by gel electrophoresis.

A
  1. Mass/number of amino acids/polypeptides; 2. Charge;

3. R groups (differ);

36
Q

One hypothesis for LP in humans suggests that the selective pressure was related to some human populations farming cattle as a source of milk.
Describe how farming cattle as a source of milk could have led to an increase in LP.

A

09.1
1. LP due to mutation
OR
Allele due to mutation;
2. Milk provides named nutrient;
3. Individuals with LP more likely to survive and reproduce
OR
Individuals with advantageous allele more likely to survive and reproduce;
4. Directional selection;
5. Frequency of allele increases (in the offspring/next generation);

37
Q

Use the information provided to explain why the number of people showing LP would
rapidly increase once selection for this condition had been established.

A

1.Dominant allele;
2.(Always) expressed/shown (when present in phenotype/offspring)
OR
Expressed when only one (dominant allele) present;

38
Q

Lactase persistence is caused by a mutation in DNA. This mutation does not occur in the gene coding for lactase.
Suggest and explain how this mutation causes LP.

A

Mutation in promoter (DNA/gene) for transcription factor
OR
Mutation in promoter (region/DNA) for the gene
OR
Mutation in gene for transcription factor;
2. Lactase gene continues to be transcribed/active

39
Q

People with complete achromatopsia have difficulty in seeing detail (lines 2–3).
Explain why.

A

No (functional) cones
OR
Only rods;
2. Cones are connected to a single neurone
OR
Several rods connected to a single neurone;
3. (Cones) Separate (sets of) impulses to brain
OR
(Rods) Single (set of) impulse/s to brain;

40
Q

Use the Hardy-Weinberg equation to calculate the percentage of this population who are heterozygous for this disorder.
Show your working.

A

Correct answer in range 42 – 44% = 2 marks;

41
Q

Red-green colour blindness affects more men than women (lines 7–8).
Explain why.

A
  1. (Gene/allele) is on the X chromosome;
  2. Females require two alleles/females can be heterozygous/carriers and males require one
    allele;
42
Q

People with red-green colour blindness are unable to distinguish between red and green, and also between other colours (lines 8–10).
Explain why.

A
  1. Green sensitive pigment/cones non-functional
    OR
    Cones that detect green light non-functional; 2. Three different types of pigment/cone;
  2. Other/different colours (‘seen’) due to stimulation of more than one cone/pigment;
43
Q

Current research into the treatment of red-green colour blindness involves the use of induced pluripotent stem cells (iPS cells) (lines 17–19).
Suggest how iPS cells could correct red-green colour blindness.

A
  1. (iPS cells) divide;

2. (iPS cells) develop/differentiate into (green sensitive) cones;

44
Q

The use of iPS cells could have advantages over the use of gene therapy to correct red-green colour blindness (lines 19–20).
Using the information from the passage, suggest and explain reasons why.

A

1 (Use of iPS cells) long-term;

  1. (Use of iPS cells) less chance of rejection/immune response;
  2. (Use of iPS cells) single treatment;
  3. Harm/side effects from using viruses (in gene therapy);