Question
1 (a) The squirrel monkey, Saimiri sciureus, of Costa Rica has become an endangered species.
Fig. 1.1 shows a squirrel monkey.
Explain what is meant by the term endangered species.
……………………………………………………………………………………………………………………..[2]
(b) Discuss possible ways in which the squirrel monkey could be protected.
……………………………………………………………………………………………………………………..[4] [Total: 6]
▶️Answer/Explanation
Ans:
(a) $1 \quad$ species threatened with extinction ;
2 numbers reduced to critical level / population too small ;
3 such low numbers that reproduction is affected ;[2max]
(b) $1 \quad$ (maintain colony) in zoo ;
2 captive breeding (programme);
3 assisted reproduction ; e.g. IVF
4 educate public ;
5 national parks / conservation areas ;
6 habitat protection ;
7 ban, hunting / poaching ;[4max]
Question
2 (a) Asellus aquaticus is a small freshwater crustacean.
200 A. aquaticus were released into a pond where there had previously been none. The pond was favourable for their growth and reproduction.
Describe and explain the expected changes in the population size of A. aquaticus over the following few months.
……………………………………………………………………………………………………………………..[5]
(b) In order for natural selection to occur a population must show phenotypic variation.
Explain why variation is important in natural selection.
……………………………………………………………………………………………………………………..[2] [Total: 7]
▶️Answer/Explanation
Ans:
2 (because) adjusting to pond environment ;
(then) steep increase / log phase / exponential increase / rapid growth or reproduction phase ;
(because) abundant food source / named other factor ;
stationary phase ;
fall in population size / death phase / decline phase ;
(due to) predation / build up of waste ;
8 competition for named resource ; e.g. food shortage
9 idea of further increase and fall / ref. population size may be cyclic ;[5max]
(b) variation means the presence of different characteristics ;
resulting in different survival rates / AW ;
(leads to) reproductive, success / failure ; [2 max]
Question
3 Proteases that work in alkaline conditions are made in large quantities for use in the detergent industry. The microorganism that is generally used for this is the bacterium Bacillus subtilis. An investigation was carried out to compare three potential production methods:
● using free cells of B. subtilis
● using B. subtilis cells immobilised in cubes of agar
● using B. subtilis cells immobilised in beads of sodium alginate.
To immobilise the cells in agar, the agar was dissolved and cooled. A suspension of B. subtilis was then added. The agar-bacterium mixture was poured into sterile dishes and allowed to solidify. It was then cut into cubes with sides of 2 mm.
(a) (i) Explain why the agar was cooled before the suspension of B. subtilis was added.
………………………………………………………………………………………………………………[1]
(ii) Describe how cells of B. subtilis could be immobilised in beads of alginate.
………………………………………………………………………………………………………………[3]
(b) A liquid medium containing glucose, a nitrogen source and various mineral ions was made up, and $50 cm_{3}$ placed into each of three flasks.
Samples of a culture of free cells of B. subtilis, agar cubes containing immobilised B. subtilis and alginate beads containing B. subtilis were placed in the three flasks. Each flask contained the same number of bacteria. All the flasks were incubated at 37 °C for 48 hours.
Samples of the liquid medium in each flask were taken at six hourly intervals and the concentration of protease measured.
The results are shown in Fig. 3.1.
(i) With reference to Fig. 3.1, compare the results for the free cells of B. subtilis and cells immobilised in alginate beads.
………………………………………………………………………………………………………………[4]
(ii) Suggest why lower concentrations of protease were produced by B. subtilis
immobilised in agar cubes than B. subtilis immobilised in alginate beads.
………………………………………………………………………………………………………………[2]
(c) Two new cultures of immobilised B. subtilis were set up as described in (b). However, this time a repeat batch fermentation method was used, in which the liquid medium was replaced every 24 hours. This was continued until the cubes or beads had begun to disintegrate.
The results are shown in Table 3.1.
With reference to Table 3.1
(i) calculate the percentage increase in the total protease produced when the bacteria were immobilised in alginate rather than agar.
Show your working. ……………………………………………[2]
(ii) explain why using bacteria immobilised in alginate rather than agar would be a more cost-effective production of protease.
………………………………………………………………………………………………………………[3]
▶️Answer/Explanation
Ans:
(a) (i) so that, the bacteria were not killed / enzymes not denatured ;
(ii) 1. bacteria put into (solution of) sodium alginate ;
2. place mixture in syringe ;
3. add drops of mixture to calcium chloride solution ;
4. calcium ions replace sodium ions (to form beads) ;
5. bacteria trapped in beads ;[3max]
(b) (i) note comparison between blue line and black line ignore references to red line – agar
1. both increase up to, $18 / 24$, hours ;
2. both similar, initially / up to 18 hours ;
3. biggest difference at 24 hours / rate of increase for immobilised cells greater than free cells between 18 and 24 hours ;
4. after 24 hours immobilised cells rate decreases while free cells rate continues to increase or after 39 hours free cells rate is greater than immobilised cells rate ;
5. free cells final concentration is still lower than highest value attained by immobilised cells ;
6. use of comparative figures ;[4max]
(ii) 1. (could be) less surface area (to volume ratio) in cubes than beads ;
2. (could be) a greater diffusion distance to centre of cubes than beads ;
3. agar may be less permeable (to substrate) than alginate ;
4. something in agar may inhibit bacterial enzymes ;
5. some protease adsorbed by agar ; [2 max]
(c) $\quad$ (i) $82.14 / 82.1 / 82(\%)$; ;
allow one mark for suitable working if incorrect answer
(ii) 1. can use alginate (beads) many times ;
2. (reduces cost of), materials / energy / labour ;
3. fewer bacterial cultures needed / less time spent immobilising bacteria ;
4. more protease produced (per hour) (using alginate) ;
5. can run fermentation for longer time ;
6. less time wasted between fermentations ;
answers must imply comparison [3max]
Question
4 Modern varieties of wheat have developed from numerous hybridisation events between different species of wild grasses. Fig. 4.1 shows some of the possible steps that are believed to have been involved in the development of bread wheat, Triticum aestivum.
The letters A, B and C represent three different sets of seven chromosomes.
(a) Complete Fig. 4.1 by writing letters to represent the sets of chromosomes in bread wheat.
Write your answer on Fig. 4.1. [1]
(b) Explain why hybridisation between emmer wheat and goat grass 2 would have produced a sterile hybrid, if doubling of chromosome number had not occurred.
……………………………………………………………………………………………………………………..[3]
(c) With reference to Fig. 4.1, suggest why Triticum urartu and Triticum turgidum are classified as different species.
……………………………………………………………………………………………………………………..[2]
(d) Triticum turgidum emerged as a new species without being geographically isolated from Triticum urartu.
Outline how geographical isolation may result in speciation.
……………………………………………………………………………………………………………………..[3] [Total: 9]
▶️Answer/Explanation
Ans:
(a) $\quad$ AABBCC ;[1]
(b) if doubling of chromosomes has not occurred
1 chromosomes would not be able to pair ;
2 because chromosomes in the two sets are not homologous ;
3 during, prophase $1 /$ meiosis 1 ;
4 (therefore) gametes cannot be produced ;[3max]
(c)
1 unable to, breed / reproduce ;
2 to produce fertile offspring ;
3 reproductively isolated ;[2max]
(d) $1 \quad$ species split into two populations by (geographical) barrier ;
2 different, selection pressures / (environmental) conditions, (on the two populations) ;
3 different features, selected / advantageous ;
4 change in, gene pools / allele frequencies ;
$5 \quad$ (over time) become unable to interbreed ;[3max]
Question
5 (a) Hormones are secreted by endocrine glands.
Explain what is meant by the term endocrine gland.
……………………………………………………………………………………………………………………..[2]
(b) Fig. 5.1 shows the changes in concentration in the blood of follicle stimulating hormone (FSH) and luteinising hormone (LH) during the first half of the menstrual cycle.
With reference to Fig. 5.1, describe,
(i) the changes that take place in the ovary during this time, as a result of the action of FSH
………………………………………………………………………………………………………………[2]
(ii) the role of LH.
………………………………………………………………………………………………………………[1]
(c) In preparation for in-vitro fertilisation (IVF), women are injected with FSH. Explain why treatment with FSH is a necessary preparation for IVF.
……………………………………………………………………………………………………………………..[2]
(d) The standard treatment with FSH and clomiphene (clomifene) causes significant side-effects. Clomiphene occupies oestrogen receptors, blocking a negative feedback mechanism.
(i) Explain briefly what is meant by negative feedback.
………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………[1]
(ii) Outline the feedback mechanism that is blocked by clomiphene.
………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………[1]
(e) Recently a so-called ‘mild’ treatment has been introduced in the hope of avoiding the side-effects of the standard treatment. This treatment does not use clomiphene. Instead, an antagonist to LH secretion is used.
The days in the first half of the menstrual cycle on which injections of FSH and clomiphene are given in the two treatments are shown by asterisks (*) in Fig. 5.2.
(i) With reference to the concentrations of LH shown in Fig. 5.1, show, using an asterisk on Fig. 5.2 when the antagonist to LH secretion should first be given.
Put your asterisk into the grey area on Fig. 5.2. [1]
(ii) Suggest why an antagonist to LH secretion forms part of the mild treatment.
……………………………………………………………………………………………………………….[1]
(f) The average dose of FSH given in the mild treatment is 1300 international units (IU), compared with an average dose of 1800 IU in the standard treatment. This could lead to the mild treatment being less effective.
The outcomes of an investigation into the two treatments are shown in Table 5.1.
With reference to Table 5.1, compare the effectiveness of the two treatments.
……………………………………………………………………………………………………………………..[3]
(g) FSH consists of two polypeptide chains which are encoded by genes on different chromosomes. The two genes, together with their promoters, have been inserted into bacteria to produce the hormone used in fertility treatments.
Explain briefly why promoters need to be transferred into the recipient bacteria together with the two genes for the FSH polypeptides.
……………………………………………………………………………………………………………………..[2] [Total: 16]
▶️Answer/Explanation
Ans:
(a) ductless gland ;
secretes (hormone) into blood ;[2]
(b) (i) 1. follicle, develops / matures / grows ;
2. detail follicle ; e.g. antrum / corona / theca
3. (follicle) secretes oestrogen (and progesterone) ; [2 max]
(ii) trigger ovulation / description ; [1]
c(1) to produce many (mature) oocytes at same time ;
2 superovulation;
3 make harvesting easier ;
5 IVF procedure has low success rate;[2max]
(d) (i) a change sets off events that counteract the change / AW / example described ; [1]
(ii) oestrogen inhibition of, GnRH / FSH ; [1]
(f) 1 very little difference in percentage of pregnancies resulting in live birth ;
2 standard (slightly) more oocytes (per cycle) ; ora
3 standard (slightly) more embryos (per cycle) ; ora
4 very little difference in percentage of pregnancies resulting in live
birth ;
standard (slightly) more oocytes (per cycle) ; ora
standard (slightly) more embryos (per cycle) ; ora
comparative figs ; [3 max]
(g) 1 (promoter needed) to ensure genes are, expressed / switched on ;
2 to produce, correct product / correct hormone / FSH ;
3 ref. human / eukaryote, gene in, bacteria / prokaryote ; $[2 \max ]$
Question
6 (a) A husband and wife who already have a child with cystic fibrosis (CF) elected to have their second child tested for the condition while still a fetus in very early pregnancy. The results of the test, a DNA banding pattern, were discussed with a genetic counsellor.
The relevant DNA banding pattern produced by electrophoresis is shown in Fig. 6.1.
With reference to Fig. 6.1, explain why,
(i) the fetus will develop CF,
………………………………………………………………………………………………………………[1]
(ii) the positions of the bands of DNA of the first child and of the fetus indicate that the mutant allele for CF has a deletion in comparison with the normal allele.
………………………………………………………………………………………………………………[2]
(b) Explain briefly the need to discuss the result of the test with a genetic counsellor.
……………………………………………………………………………………………………………………..[4] [Total: 7]
▶️Answer/Explanation
Ans:
(a) (i) same band of DNA as, first / affected, child ; [1]
(ii) 1. father and mother, have normal and mutant alleles / are heterozygous ;
2. mutant / CF, DNA is, shorter / lighter ;
3. therefore travels further ; [2 max]
(b) $1-1$ outcome of test needs explanation / counsellor gives advice on options ;
2 already have one affected child to care for or problems / cost, of care ;
3 ref. termination ;
4 life expectancy increasing with improved drugs ;
5 gene therapy, not as yet successful / likely to be temporary ;
6 possibility of, pre-implantation genetic diagnosis (PGD) / artificial insemination by donor sperm (AID), on another occasion ; $[4 \max ]$
Question
7 (a) The fruit fly, Drosophila melanogaster, feeds on sugars found in damaged fruits. A fly with normal features is called a wild type. It has a striped body and its wings are longer than its abdomen. There are mutant variations such as an ebony coloured body or
vestigial wings. These three types of fly are shown in Fig. 7.1.
Wild type features are coded for by dominant alleles, A for wild type body and B for wild type wings.
Explain what is meant by the terms allele and dominant.
allele ………………………………………………………………………………………………………………..
dominant …………………………………………………………………………………………………………..[2]
(b) Two wild type fruit flies were crossed. Each had alleles A and B and carried alleles for ebony body and vestigial wings.
Draw a genetic diagram to show the possible offspring of this cross.
(c) When the two heterozygous fruit flies in (b) were crossed, 384 eggs hatched and developed into adult flies.
A chi-squared $\left(\chi^2\right)$ test was carried out to test the significance of the differences between observed and expected results.
$
\begin{aligned}
& \chi^2=\Sigma \frac{(\mathrm{O}-\mathrm{E})^2}{E} \\
& \text { where } \Sigma=\text { sum of } \\
& \mathrm{O}=\text { observed value } \\
& \mathrm{E}=\text { expected value } \\
&
\end{aligned}
$
(i) Complete the missing values in Table 7.1.
(ii) Calculate the value for $\chi^2$.
$
\chi^2=
$
Table 7.2 relates $\chi^2$ values to probability values.
As four classes of data were counted the number of degrees of freedom was $4-1=3$. Table 7.2 gives values of $\chi^2$ where there are three degrees of freedom.
(iii) Using your value for $\chi^2$, and Table 7.2, explain whether or not the observed results were significantly different from the expected results.
………………………………………………………………………………………………………………[2] [Total: 14]
▶️Answer/Explanation
Ans:
(a) allele
different / alternative, form of a gene ; A variety of a gene dominant
(allele) that always expresses itself in the phenotype when present /
(allele) which influences the phenotype even in the presence of an
alternative allele / AW ; [2]
(b)
parental phenotype; e.g. striped / long $x$ striped / long A wild wild
parental genotype; e.g. AaBb x AaBb
gametes; e.g. $\mathrm{AB} \mathrm{Ab} \mathrm{aB} \mathrm{ab}$
offspring genotypes $; ;$
offspring phenotypes; must be linked to genotypes
accept other symbols if key used
penalise once for no key but only if genetic cross works [2]
(b)
parental phenotype; e.g. striped / long $x$ striped / long A wild $\times$ wild
parental genotype; e.g. AaBb x AaBb
gametes ; e.g. $A B A b$ aB ab
offspring genotypes $; ;$
offspring phenotypes; must be linked to genotypes
accept other symbols if key used
penalise once for no key but only if genetic cross works[6]
(c) (i)
[3]
(ii) 2.78 ; apply ecf [1]
(iii) $\chi^2$ value represents probability of $>0.05$; no significant difference ;
(probability shows) differences due to chance ; [2max]
Question
8 (a) Fig. 8.1 shows the results from two experiments carried out to investigate the effect of light intensity and carbon dioxide concentration on the rate of photosynthesis.
(i) Describe and explain the results shown in Fig. 8.1 for experiment 1.
………………………………………………………………………………………………………………[3]
(ii) Describe and explain the difference between the results for experiment 1 and experiment 2.
………………………………………………………………………………………………………………[3]
(b) The optimum temperature for many plants living in temperate regions is approximately $25^{\circ} \mathrm{C}$.
Explain why the rate of photosynthesis in these plants decreases at temperatures above $25^{\circ} \mathrm{C}$.
……………………………………………………………………………………………………………………..[5] [Total: 11]
▶️Answer/Explanation
Ans:
(a) (i) at low light intensity
1. rate of photosynthesis increases as light intensity increases ;
2. light intensity is limiting factor ;
at higher light intensity
3. graph, levels off / forms a plateau / rate becomes constant ;
4. $\mathrm{CO}_2$ / some other factor, becomes limiting ;[3max]
(ii) 1. above light intensity of 1 rate is always higher for expt. 2 ;
2. plateau reached at lower light intensity for expt. 1 ;
3. maximum / plateau, rate is double for expt. 2 ;
4. expt 2 has much more $\mathrm{CO}_2$ (conc) (compared to expt 1 );
5. $\mathrm{CO}_2$, no longer limiting after 4.2 in expt. 2 / is limiting in expt. 1 up to 2.8 ;[3max]
(b)
1 enzymes, denatured / active site changes shape ;
2 rubisco / enzyme in cyclic photophosphorylation;
3 Calvin cycle affected / description ;
4 less photolysis ;
5 less ATP produced ;
6 increased rate of respiration ;
7 respiration rate faster than photosynthesis rate / ref. compensation point ;
8 increased rate of transpiration ;
9 stomatal closure ;
less $\mathrm{CO}_2$ uptake ;[5max]
Question
Section B
Answer one question.
9 (a) Describe the process of glycolysis. [7]
(b) Describe the structure and synthesis of ATP and its universal role as the energy currency in all living organisms. [8] [Total: 15]
10 (a) Describe a reflex arc and explain why such reflex arcs are important. [7]
(b) Describe the structure of a myelin sheath and explain its role in the speed of transmission of a nerve impulse. [8] [Total: 15]
▶️Answer/Explanation
Ans:
(a) $11 \quad$ (glucose) phosphorylated by ATP ;
raises energy level / overcomes activation energy ;
hexose bisphosphate ;
lysis / splitting, of, glucose / hexose ; $\quad \mathbf{R}$ sugar splitting
breaks down to two TP ; A GALP / GADP / G3P / PGAL
$6 \mathrm{C} \rightarrow 2 \times 3 \mathrm{C}$
dehydrogenation / description ;
$\underline{2}$ NAD reduced formed (from each TP to pyruvate formed) ;
4 ATP produced / net gain of 2 ATP ;
pyruvate produced ;
reduced NAD $\rightarrow$ oxidative phosphorylation / redox ;
accept flow diagram
(b)
12. nucleotide ;
13. adenine + ribose / pentose + three phosphates ;
14. loss of phosphate leads to energy release / hydrolysis releases $30.5 \mathrm{~kJ}$
15. $\mathrm{ADP}+\mathrm{Pi} \leftrightarrow \mathrm{ATP}$ (reversible reaction)
16.synthesised during, glycolysis / Krebs cycle / substrate level phosphorylation ;
17.synthesised, using electron carriers / oxidative phosphorylation / photophosphorylation ;
18.in, mitochondria / chloroplasts ;
19.ATP synthase / ATP synthetase ;
20.chemiosmosis / description;
21.used by cells as immediate energy donor ;
22.link between energy yielding and energy requiring reactions / AW ;
23.active transport / muscle contraction / Calvin cycle / protein synthesis ;