New WJEC GCE Biology Unit 2 2400U20-1

WJEC 2016 Online Exam Review

New WJEC Biology Unit 2 2400U20-1

All Candidates' performance across questions

Question Title N Mean S D Max Mark F F Attempt %1 3609 5.6 2.6 14 39.8 99.92 3608 2.4 1.6 8 29.8 99.93 3611 5.8 2.5 14 41.5 1004 3606 5.6 2.9 14 40 99.85 3602 3.6 2 8 44.5 99.76 3588 3.1 2 13 23.5 99.37 3461 2.8 1.8 9 31.3 95.8

39.8

29.8

41.5

40

44.5

23.5

31.3

0 10 20 30 40 50 60 70 80 90 100

1

2

3

4

5

6

7

Facility Factor %

Que

stio

n

New WJEC Biology Unit 2 2400U20-1

Sticky Note

Usually the question number

Sticky Note

The number of candidates attempting that question

Sticky Note

The mean score is calculated by adding up the individual candidate scores and dividing by the total number of candidates. If all candidates perform well on a particular item, the mean score will be close to the maximum mark. Conversely, if candidates as a whole perform poorly on the item there will be a large difference between the mean score and the maximum mark. A simple comparison of the mean marks will identify those items that contribute significantly to the overall performance of the candidates. However, because the maximum mark may not be the same for each item, a comparison of the means provides only a partial indication of candidate performance. Equal means does not necessarily imply equal performance. For questions with different maximum marks, the facility factor should be used to compare performance.

Sticky Note

The standard deviation measures the spread of the data about the mean score. The larger the standard deviation is, the more dispersed (or less consistent) the candidate performances are for that item. An increase in the standard deviation points to increased diversity amongst candidates, or to a more discriminating paper, as the marks are more dispersed about the centre. By contrast a decrease in the standard deviation would suggest more homogeneity amongst the candidates, or a less discriminating paper, as candidate marks are more clustered about the centre.

Sticky Note

This is the maximum mark for a particular question

Sticky Note

The facility factor for an item expresses the mean mark as a percentage of the maximum mark (Max. Mark) and is a measure of the accessibility of the item. If the mean mark obtained by candidates is close to the maximum mark, the facility factor will be close to 100 per cent and the item would be considered to be very accessible. If on the other hand the mean mark is low when compared with the maximum score, the facility factor will be small and the item considered less accessible to candidates.

Sticky Note

For each item the table shows the number (N) and percentage of candidates who attempted the question. When comparing items on this measure it is important to consider the order in which the items appear on the paper. If the total time available for a paper is limited, there is the possibility of some candidates running out of time. This may result in those items towards the end of the paper having a deflated figure on this measure. If the time allocated to the paper is not considered to be a significant factor, a low percentage may indicate issues of accessibility. Where candidates have a choice of question the statistics evidence candidate preferences, but will also be influenced by the teaching policy within centres.

Sticky Note

Oxygen supplied directly to respiring cells is a correct advantage of the tracheal system and limitation of size of an insect is correct as a disadvantage. 2 marks

Sticky Note

It is correctly stated that an amoeba has a large surface area to volume ratio and that the diffusion distance is short. The candidate correctly implies that the diffusion of gases is sufficient to supply demand. 3 marks

Sticky Note

6000 : 1 is incorrect. 0 marks

Sticky Note

The candidate appreciates that the dog has a higher metabolic rate than a fish but no reference is made to less energy being required by a fish because less energy is required for support. 1 mark

Sticky Note

The circulatory system and ventilation movements are both stated as being involved in maintaining the concentration gradient. 2 marks

Sticky Note

A very clear and concise response. The candidate clearly states that mammals evolved when oxygen levels were lower. The candidate correctly states that the biconcave shape increases the surface area and decreases the diffusion distance. The lack of a nucleus increasing the amount of haemoglobin in the RBC is also stated. 4 marks awarded.

Sticky Note

Total = 12

7

Question Marking details Marks available

AO1 AO2 AO3 Total Maths Prac

4 (a) i Some water used by {photosynthesis/metabolic reactions}/

water produced in respiration/

measures the rate of absorption not the rate of transpiration

1 1

ii Any 3 (x1) from:

humidity/ or description (1)

wind/ air currents (1)

surface area of leaves (1)

age of leaves (1)

Accept air pressure

NOT same number/ mass of leaves/ length of stem/ plant

2 1 3 3

(b) Lower surface of oak leaf shaded/ or description of/ ORA (1)

so higher density of stomata to reduce water loss (1)

OR

neither surface of wheat shaded/ or description of (1)

equal distribution of stomata water loss equal both sides (1)

2 2

© WJEC CBAC Ltd. 8



(c) 1. (Potassium ions/ malate) reduce water potential in (guard) cell (1)

Accept osmotic pressure increases/ osmotic potential decreases/

solute potential decreases/ hypertonic to outside

2. water moves in by osmosis (down water potential gradient) (1)

3. (Turgor) pressure inside (guard) cell increases/ cells become turgid

(1) NOT cells expand

4. ends of guard cell have a thinner wall than centre/ ORA (1)

5. ends of guard cell expand and stomata opens (1)

3 2 5

(d) i 2 x π x 2 x 423 = 5312.9 (1)

= 5310 ( to 3 sig figs) (1)

Allow 5320 if they use value of π from calculator.

5310/ 5320 = 2 marks

5312.9/ 5313/ 5315.6/ 5316 = 1 mark

Evidence of 2 πr x 423 = 1 mark

2 2 2

ii (water molecules) escape more readily from) species B because it

has larger (total) circumference.

Ecf if calculation incorrect in (i)

1 1

Question 4 total 5 7 2 14 2 3

Sticky Note

Correct reference to protein digestion in stomach but hydrolysis of proteins into polypeptides incorrect.

Sticky Note

There is an explanation of protein digestion and although incorrect an attempt has been made to account for the higher level of protein required by a horse and to explain why there are higher levels of nitrogen in the faeces. This is sufficient to put it in the mid band. The essay was awarded 4 marks.

Sticky Note

The candidate appreciates that mutualistic bacteria produce cellulases which break down cellulose but no reference is made to the site of this activity in the cow and the horse. The link between this to the increased protein required in the horses diet and a higher level of organic nitrogen in the faeces is not made.

Sticky Note

Correct reference to absorption of amino acids from small intestine into blood.

Sticky Note

Correct reference to presence of endopeptidases and exopeptidases in small intestine but where theses enzymes act has been omitted.

3

Sticky Note

Correctly stated that an amoeba has a large surface area to volume ratio but the significance of this is not given and no reference to diffusion distances is given. 1 mark

Sticky Note

Candidate gives a correct advantage of the tracheal system but disadvantage is incorrect. 1 mark

Sticky Note

Correct surface area : volume ratio given. 1 mark

Sticky Note

Correctly stated that fish have a lower metabolic rate but the reason for this is not given. 1 mark

Sticky Note

The role of the circulatory system in the maintenance of a concentration gradient is given but there is no reference to ventilation. 1 mark

Sticky Note

The candidate correctly states that mammals evolved when there was less oxygen in the atmosphere and that their red blood cells have adapted by having a short diffusion path and no nucleus for more haemoglobin. 3 marks

Sticky Note

Total = 8 marks

(2400U20-1) Turn over.

24

00

U2

01

11

11

11Examiner

only

© WJEC CBAC Ltd.

4. All terrestrial plants lose water from their leaves by a process called transpiration.

(a) A student used a potometer to estimate the transpiration rate of a leafy shoot over a twenty-four hour period.

(i) Explain why the water uptake from the potometer gives an estimate of the transpiration rate and not its true value. [1]

(ii) Give three other variables that would need to be controlled when using a potometer to investigate the difference in transpiration rates of leafy shoots of two different plant species at the same light intensity and temperature. [3]

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(b) The table shows the average density and distribution of stomata on the leaves of oak and wheat.

Wheat (Triticum aestivum) Oak (Quercus robur)

PlantDensity of Stomata (number mm–2)

Upper leaf surface Lower leaf surfaceWheat 50 40Oak 0 340

Explain the distribution of stomata in wheat and oak. [2]

12

(2400U20-1)12

Examineronly

© WJEC CBAC Ltd.

(c) The diagrams below show the arrangement of guard cells in a leaf of wheat under two different environmental conditions.

Thin cellulose cell wall

Cytoplasm

Thick cellulose cell wallStoma

Guard cell

Stomata closed Stomata open

Stomata open when potassium ions are actively transported into the guard cells. Explain how this would cause wheat stomata to open. [5]


24

00

U2

01

13

13

13Examiner

only

© WJEC CBAC Ltd.

(d) Water is lost through stomata by transpiration. This results in a shell of water vapour forming over the stomatal pores as shown.

spongy mesophyll of leaf

epidermis

diffusion shell – area of high level of water vapour

Air

stomatal pore

Water molecules from the edges of stomatal pores evaporate more readily into the air.

The stomatal density and the diameter of the stomatal pores were recorded for the leaves of two species of plant. The stomatal pores were circular and the total area of the pores was approximately the same in both plants.

Plant species Density of stomata (number mm–2)

Mean diameter of stomatal pore

(µm)

Total circumference of stomata(µm mm–2)

A 110 8 2 760

B 423 4

Circumference of a circle = 2πr r = radius of circle π = 3.14

(i) Calculate the total circumference of the stomatal pores for species B. Give your answer to three significant figures. [2]

Circumference = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(ii) Use these figures to identify which plant species, A or B, has the higher transpiration rate. Explain your answer. [1]

14

Sticky Note

Correct statement. 1 mark

Sticky Note

Three correct variables. 3 marks

Sticky Note

Correct reference to no stomata on upper surface in presence of high light intensities to reduce water loss. 2 marks

Sticky Note

Correct reference to a decrease in water potential, water moving in by osmosis and the cells becoming turgid. However the candidate failed to apply their knowledge to this shape of guard cells and so did not access the last two marks points. 3 marks

Sticky Note

Calculation correct. 2 marks

Sticky Note

Correct reference to larger total circumference. 1 mark

Sticky Note

Total = 12 marks

2

(2400U20-1)02

Examineronly

© WJEC CBAC Ltd.

Answer all questions.

1. (a) The table below shows the relationship between length, surface area and volume for three different organisms.

Organism Length Surface area (m2)

Volume (m3)

Surface area to volume ratio

amoeba 1µm 6 × 10–12 1 × 10–18 6000000 : 1

housefly 10mm 6 × 10–8 1 × 10–12

dog 1m 6 × 100 1 × 100 6 : 1

The housefly is an organism adapted to a terrestrial mode of life, possessing an internal tracheal system for gas exchange.

(i) Calculate the surface area to volume ratio of the housefly. [1]

Surface area to volume ratio = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(ii) Give one advantage and one disadvantage of the tracheal system. [2]

Advantage: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Disadvantage: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(b) Use the table to explain why an amoeba does not need a specialised gas exchange surface or circulatory system. [3]


24

00

U2

01

03

03

3Examiner

only (c) State how mammals maintain the concentration gradient at their gas exchange surface. [2]

(d) Suggest why cellular demand for oxygen is lower in a fish compared to a dog of the same size. [2]

© WJEC CBAC Ltd.

4

(2400U20-1)04

Examineronly

© WJEC CBAC Ltd.

How do you account for the difference in size and structure of bird and mammalian red blood cells? [4]

Red blood cells from birds

14

(e) Mammals evolved in the Triassic period when levels of oxygen in the atmosphere were 50% lower than now. Birds evolved much later, in the Jurassic period, when oxygen levels in the atmosphere approached present day levels.

The red blood cells of birds are elliptical in shape and have a nucleus.

Mammalian red blood cells are biconcave in shape, do not have a nucleus and are much smaller than the red blood cells of birds.

Sticky Note

Response not worthy of credit. Answer required reference to number of stomata linked to water loss and whether the leaf was shaded. 0 marks

Sticky Note

Candidate states that the same species of plant should be used, in the stem of the question it clearly states 'two different species' Credit was not given for same size of leaf, we needed same surface area. 0 marks

Sticky Note

Correct statement. 1 mark

Sticky Note

Mark awarded for decrease in water potential in guard cells but there is no reference to water moving in by osmosis. It is stated that the turgor pressure increases but there is no reference to the thin walled ends of the guard cells or to the expansion of the ends leading to the opening of the stomata. 2 marks

Sticky Note


Sticky Note

Correct reference to greater total circumference. 1 mark

Sticky Note

Total = 6

Sticky Note

Correct reference to the digestion of proteins beginning in the abomasum (true stomach) with a reference to endopeptidase breaking protein into tri and dipeptides.

Sticky Note

The essay addresses all aspects of the question and is placed in the top band. There is insufficient detail of protein digestion for 9 marks and 8 marks were awarded.

Sticky Note

Correct reference to the digestion of cellulose by bacteria taking place in the caecum of the horse which is after the area of the gut in which digestion and absorption of protein takes place.

Sticky Note

Reference to digestion of protein using exopeptidases (but no reference as to site of action) and absorption in the small intestine.

Sticky Note

Candidate appreciates that a large proportion of the cows protein comes from bacteria and that bacteria use urea to make the protein.

Sticky Note

Candidate states that this is the reason why horses need a diet rich in protein and suggests that this is a reason why horse manure has a higher protein content.

© WJEC CBAC Ltd. 2

Question Marking Details Marks available


1 (a) i 60 000 : 1

Accept correct ratio not to 1 e.g. 120 000:2

1 1 1

ii Advantage:

Any one from:

Reduces water loss/ allows them to live in arid conditions (1)

No {blood / circulatory system/ pigment} required (1)

Oxygen supplied directly to the cells/ muscles (1)

Tracheoles go directly into cells/ tissues (1)

Disadvantage:

Size/ shape limitation (1)

2 2

(b) Any 3 (x1) from:

Diffusion of gases related to Surface Area (1)

Oxygen use related to volume (1)

Amoeba has large Surface Area : volume ratio (1)

Therefore diffusion of gases sufficient (to supply demand) (1)

Short diffusion distance (1)

3 3

(c) Ventilation (movements/system)/ description of replacing oxygen in

alveoli/ ORA for carbon dioxide (1)

{Blood/ transport system} (takes oxygen away from respiratory surface)/

ORA for carbon dioxide/ or description of (1)

2 2

3

Question Marking Details Marks available


(d) Any two (x1) from:

Low body temperature (1)

Metabolic rate low / example of (metabolic) reactions not

required/ working at lower rate(1)

Less energy required for support / fish are buoyant (1)

ORA for dog

Reject dogs move more than fish/ dogs are more active than fish/ more

oxygen needed for respiration

2 2

(e) 1. Mammalian (red blood cells) evolved when O2 levels were lower/

ORA (1)

2. {Biconcave/shape} increases surface area (for increased O2

absorption) (1)

3. No nucleus + can carry more haemoglobin (so increased O2

transport) / ORA(1)

4. Thin centres/small/ biconcave so short diffusion distance (so

faster diffusion) (1)

5. Small(er) in size so have a higher sa : vol ratio (1)

6. Small(er), {so more of them/ total surface area larger} (1)

Accept ref. to smaller capillaries qualified/ biconcave shape

gives flexibility to fit through capillaries

4 4


Sticky Note

Candidate clearly states that lower surface of leaf exposed to less light and therefore by having stomata on lower surface, transpiration rate is reduced. 2 marks

Sticky Note

Marks awarded for same wind intensity and same humidity. There is no reference to same surface area or same age of leaves. 2 marks

Sticky Note

No mark awarded, incorrect statement. 0 marks

Sticky Note

The marking points for the reduction of water potential leading to water moving into the guard cells by osmosis causing an increased turgor pressure are all accessed. The candidate does not relate this increase in turgor to the opening of the particular stomata shown in the diagram. 3 marks

Sticky Note


Sticky Note

Reference to larger total circumference required for the mark. 0 marks

Sticky Note

Total = 9

3

Sticky Note

It is correctly stated that an amoeba has a large surface are to volume ratio but the significance of this is not given. 1 mark

Sticky Note

Prevention of water loss is correct for an advantage of the tracheal system but the disadvantage stated is incorrect. 1 mark

Sticky Note

Correct surface area to volume ratio given. 1 mark

Sticky Note

It is stated correctly that fish have a lower metabolic rate but no reason for this is given. 1 mark

Sticky Note

The candidate has referred to a ventilation mechanism, but not referred to the circulatory system. 1 mark

Sticky Note

The candidate clearly states that mammals evolved when oxygen levels were lower. The reference to birds having a larger shape is too vague to be credited. 1 mark

Sticky Note

Total = 6

Sticky Note

An excellent introduction, the roles of exopeptidases an exopeptidases are clearly stated and an accurate reference is made to absorption of amino acids.

Sticky Note

Attempts to explain the difference in the protein requirement and difference in organic nitrogen levels have been made (although incorrect) and the description of protein digestion is excellent. As a result the essay is placed in the middle band and was awarded 6 marks.

Sticky Note

The candidate does not appreciate that bacteria are required for the breakdown of cellulose and that the protein produced by the bacteria provides a source of amino acids. No reference is made to the different sites of bacterial activity or to the implication of this.

Sticky Note

Correct reference to pepsin and trypsin.

22

(2400U20-1)22

Examineronly

© WJEC CBAC Ltd.

7. All animals need a source of amino acids for growth and repair of tissues. Bacteria, found in the gut of herbivores, can break down cellulose into energy rich molecules; these can be absorbed and used by the herbivore.

Cows produce a large volume of saliva which contains urea. The urea provides a source of nitrogen that bacteria use to make proteins.

Horses need a diet which is much richer in protein than the diet required by a cow.

Manure from horses contains 3 times as much organic nitrogen as cow manure.

HORSE

Oesophagus

Stomach

Large intestine

Small intestine Caecum – breakdown of cellulose by bacteria

COW

Oesophagus 3 chambers in oesophagus – breakdown of cellulose by bacteria

True stomach

Large intestine

Small intestineCaecum

Describe how proteins are digested and made available to the muscles of the herbivore. Using the diagrams of the gut of a cow and a horse, account for the difference in the protein requirement of a horse and a cow and the difference of organic nitrogen in the manure of both animals.

[9 QER]

© WJEC CBAC Ltd. 12



7 Pepsin in stomach, hydrolyses peptide bonds breaking down polypeptides into shorter chains of amino acids.

The pancreas produces proteases such as trypsin which breaks down polypeptide chains into shorter chains.

Cells in small intestine secrete peptidases which complete the breakdown of polypeptides into amino acids.

Ref to exopeptidases and endopeptidases.

Amino acids absorbed into the blood from small intestine transported to the muscles.

Cows large numbers of bacteria are produced in the first three chambers of the ‘stomach’

make protein using urea

when the bacteria pass into the true stomach they are killed by the acid.

Proteins {in/ from} the bacteria are then digested and absorbed

Horses do not have saliva containing urea

This explains why horses need more protein in food than cows

In horses the bacteria are found in the caecum/ large intestine

Protein in these bacteria are lost in the faeces because no digestion or absorption takes place in the large intestine.

This explains why horse manure has a higher levels of organic nitrogen.

3

4

2

9

7-9 marks Detailed explanation of protein digestion Explanation of use of urea by bacteria in cow/ digestion of bacteria Explanation of increased protein in diet of horse/ nitrogen content in manure of horse The candidate constructs an articulate, integrated account, correctly linking relevant points, such as those in the indicative content, which shows sequential reasoning. The answer fully addresses the question with no irrelevant inclusions or significant omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately.

13



4-6 marks Any two from: Explanation of protein digestion Brief explanation of use of urea by bacteria in cow/ digestion of bacteria Brief explanation of increased protein in diet/ nitrogen content in manure The candidate constructs an account correctly linking some relevant points, such as those in the indicative content, showing some reasoning. The answer addresses the question with some omissions. The candidate usually uses scientific conventions and vocabulary appropriately and accurately.

1-3 marks. Any one from: Brief explanation of protein digestion Brief explanation of ruminant digestion Brief explanation of increased protein in diet/ nitrogen content in manure The candidate makes some relevant points, such as those in the indicative content, showing limited reasoning. The answer addresses the question with significant omissions. The candidate has limited use of scientific conventions and vocabulary.

0 marks The candidate does not make any attempt or give a relevant answer worthy of credit.


Documents

New WJEC GCE Biology Unit 2 2400U20-1