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GCSE · Biology

B6 Inheritance, Variation and Evolution

AQA 8461/8462/8463 · Calculator allowed · about 105 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Reproduction and inheritance rely on several key structures found in cells.

(a)Define the term 'gamete'.(1)

(b)Define the term 'chromosome'.(1)

(c)Define the term 'gene'.(1)

(d)Define the term 'allele'.(1)

(Total for Question 1 is 4 marks)

DNA is the molecule that carries genetic information in almost all living organisms.

(a)Name the shape formed by a DNA molecule.(1)

(b)State what is meant by the 'genome' of an organism.(1)

(c)State the number of chromosomes found in a normal human body (somatic) cell.(1)

(Total for Question 2 is 3 marks)

Body cells divide by mitosis, and specialised cells in the reproductive organs divide by meiosis to produce gametes.

(a)State the type of cell division used to produce gametes.(1)

(b)State the type of cell division used for the growth and repair of body cells.(1)

(c)State two ways in which the cells produced by meiosis are different from the cells produced by mitosis.(2)

(Total for Question 3 is 4 marks)

In pea plants, the allele for tall stems (T) is dominant to the allele for short stems (t). A gardener in Kent crosses two pea plants that are both heterozygous (Tt) for stem height.

(a)Complete a genetic cross diagram to show the possible gametes and offspring genotypes from this cross.(2)

Figure (to be drawn): Blank Punnett square grid provided: parents Tt x Tt, gametes T and t along each axis, for the student to complete.

(b)Using your diagram, state the genotype ratio of the offspring.(1)

(c)State the phenotype ratio of the offspring, and calculate the percentage of offspring expected to be tall.(2)

(Total for Question 4 is 5 marks)

Cystic fibrosis is a genetic disorder caused by a recessive allele, f. The dominant allele, F, does not cause cystic fibrosis. Two parents, both unaffected carriers with genotype Ff, are expecting a baby.

(a)Complete a genetic cross diagram to show the possible genotypes of their children.(3)

Figure (to be drawn): Blank Punnett square grid provided: parents Ff x Ff, for the student to complete.

(b)Calculate the probability, as a percentage, that a child of these two parents will be born with cystic fibrosis.(1)

(c)Explain why two parents who do not have cystic fibrosis can have a child with cystic fibrosis.(1)

(Total for Question 5 is 5 marks)

The sex of a human baby is determined by one pair of chromosomes. Females have the genotype XX and males have the genotype XY.

(a)Complete a genetic cross diagram to show the probability of a baby being male or female, and state this probability.(2)

Figure (to be drawn): Blank Punnett square grid provided: mother XX x father XY, for the student to complete.

(b)Explain why it is the father's gamete, and not the mother's, that determines the sex of the child.(2)

(Total for Question 6 is 4 marks)

Polydactyly is a condition in which a person is born with extra fingers or toes. It is caused by a dominant allele, D. One parent has polydactyly and is heterozygous (Dd); the other parent does not have polydactyly (dd).

(a)Complete a genetic cross diagram and use it to calculate the probability, as a percentage, that a child of these parents will have polydactyly.(2)

Figure (to be drawn): Blank Punnett square grid provided: parents Dd x dd, for the student to complete.

(b)Polydactyly does not usually reduce a person's chance of surviving or having children. Explain why this means the allele for polydactyly is not removed from the population by natural selection, even though it is not the most common form of the gene.(2)

(Total for Question 7 is 4 marks)

Embryos created by IVF can be screened for the alleles that cause serious genetic disorders, such as cystic fibrosis, before being implanted in the mother. Evaluate the use of embryonic screening to prevent genetic disorders being passed on to children.

(Total for Question 8 is 6 marks)

Variation exists between individuals of the same species.

(a)State two causes of variation between individuals of the same species.(2)

(b)Give one example of a characteristic that is mainly caused by environmental variation.(1)

(Total for Question 9 is 3 marks)

Mutations are changes to the DNA of an organism.

(a)Define the term 'mutation'.(1)

(b)Explain why most mutations do not have a noticeable effect on the phenotype of an organism.(2)

(Total for Question 10 is 3 marks)

Required practical: Two students, Amara and Josh, from a school near Bristol investigated variation in the shell banding pattern of the land snail Cepaea nemoralis. This snail's shell can be plain (unbanded) or striped (banded), a characteristic controlled by a gene. The students sampled snails from two habitats: a short grassland meadow and a hedgerow with variable light and shade. Their results are shown below. Grassland: 32 banded out of 40 snails sampled. Hedgerow: 45 banded out of 60 snails sampled.

(a)Describe how the students could use quadrats to obtain an unbiased, random sample of snails within each habitat.(2)

(b)Suggest one variable, other than the species sampled, that the students should try to keep the same when comparing the two habitats.(1)

(c)Calculate the percentage of snails that were banded in the grassland habitat.(2)

(d)Calculate the percentage of snails that were banded in the hedgerow habitat.(1)

(e)Using your answers to (c) and (d), and the idea of natural selection, explain why banded snails might be more common in the grassland than in the hedgerow.(3)

(f)Explain why increasing the sample size from 40-60 snails per habitat to 400-600 snails per habitat would make the students' conclusion more valid.(2)

(g)Suggest one reason, other than camouflage and predation, why the frequency of the banded allele might differ between two separate snail populations.(1)

(h)State what is meant by 'natural selection', using this investigation as an example.(1)

(i)Explain one limitation of using shell-banding percentage alone as evidence that natural selection is acting on this population.(2)

(Total for Question 11 is 15 marks)

Charles Darwin proposed the theory of evolution by natural selection. Describe the process of natural selection, using the example of a population of animals in which some individuals are, by chance, better camouflaged than others.

(Total for Question 12 is 4 marks)

A microbiologist in a hospital laboratory grew a culture of 5,000,000 bacteria. Before any antibiotic was added, 250 of these bacteria already carried a mutation that gave them resistance to the antibiotic meticillin.

(a)Calculate the percentage of the original bacterial population that was resistant to meticillin.(2)

(b)The bacterial culture was then treated with meticillin. Explain why, after treatment, nearly all of the surviving bacteria were resistant to the antibiotic, even though resistant bacteria made up such a small percentage of the population before treatment.(3)

(Total for Question 13 is 5 marks)

Fossils provide evidence for how life on Earth has changed over time.

(a)Describe two ways in which fossils can form.(2)

(b)Explain why the fossil record is incomplete.(2)

(Total for Question 14 is 4 marks)

For many years, living organisms were classified into a system of five kingdoms. New evidence has led scientists to propose a different system of classification.

(a)Name the three domains in the three-domain system of classification proposed by Carl Woese.(1)

(b)Explain why new evidence led scientists to divide organisms into these three domains, rather than continuing to use the five-kingdom system.(3)

(Total for Question 15 is 4 marks)

Selective breeding is used by farmers to produce plants and animals with desirable characteristics. A farmer in Norfolk wants to breed wheat plants that are more resistant to a fungal disease.

(a)Describe the process the farmer could use to selectively breed wheat plants with increased disease resistance.(4)

(b)State one advantage of selectively breeding wheat in this way.(1)

(c)State one disadvantage of selectively breeding wheat in this way.(1)

(Total for Question 16 is 6 marks)

Bacteria can be genetically engineered to produce human insulin. Describe how a bacterium is genetically engineered so that it produces human insulin.

(Total for Question 17 is 6 marks)

Genetically modified (GM) crops, such as crops engineered to be resistant to insect pests, are grown in some countries.

(a)State one advantage of growing GM crops that are resistant to insect pests.(2)

(b)State one disadvantage or concern associated with growing GM crops.(2)

(Total for Question 18 is 4 marks)

In snapdragon plants, flower colour can show codominance. The allele R codes for red pigment and the allele r codes for white pigment; a plant with genotype Rr has pink flowers, because both alleles are expressed.

(a)A homozygous red plant (RR) is crossed with a homozygous white plant (rr). Complete a genetic cross diagram to find the genotype and phenotype of all the offspring, and use your diagram to explain what is meant by 'codominance'.(3)

Figure (to be drawn): Blank Punnett square grid provided: parents RR x rr, for the student to complete.

(b)Two of the pink (Rr) offspring from part (a) are crossed together. Complete a genetic cross diagram to find the genotype and phenotype ratio of this second generation.(3)

Figure (to be drawn): Blank Punnett square grid provided: parents Rr x Rr, for the student to complete.

(Total for Question 19 is 6 marks)

A student crosses two heterozygous pea plants (Tt x Tt), where tall (T) is dominant to short (t), and grows 200 offspring from the seeds produced.

(a)Calculate the expected number of tall plants and the expected number of short plants among the 200 offspring.(2)

(b)When the student counted the actual plants, 138 were tall and 62 were short. Suggest why this observed result is different from the expected result calculated in part (a).(2)

(Total for Question 20 is 4 marks)

Mark scheme · B6 Inheritance, Variation and Evolution

Question 1

(a) B1 a sex cell (e.g. sperm or egg cell) used in sexual reproduction (oe)
(a) Answer: A sex cell, such as a sperm cell or egg cell, used in sexual reproduction.
(b) B1 a long molecule of coiled/packaged DNA (that carries genetic information) (oe)
(b) Answer: A long, coiled molecule of DNA found in the nucleus, made up of many genes.
(c) B1 a small section of DNA (on a chromosome) that codes for a particular protein/characteristic (oe)
(c) Answer: A small section of DNA on a chromosome that codes for a particular protein and so controls a characteristic.
(d) B1 one of the different (alternative) forms of a gene (oe)
(d) Answer: One of the alternative forms of a gene.

Question 2

(a) B1 double helix
(a) Answer: A double helix.
(b) B1 the entire genetic material of an organism (oe)
(b) Answer: The entire genetic material of an organism.
(c) B1 46 (23 pairs)
(c) Answer: 46 chromosomes (23 pairs).

Question 3

(a) B1 meiosis
(a) Answer: Meiosis.
(b) B1 mitosis
(b) Answer: Mitosis.
(c) B1 cells from meiosis are genetically different from each other and from the parent cell, whereas cells from mitosis are genetically identical to the parent cell and to each other
(c) B1 cells from meiosis contain half the number of chromosomes (haploid), whereas cells from mitosis contain the full number of chromosomes (diploid)
(c) Answer: Meiosis produces genetically different, haploid cells; mitosis produces genetically identical, diploid cells.

Question 4

(a) M1 correct gametes shown for each parent: T and t
(a) A1 correct offspring genotypes shown: TT, Tt, Tt, tt
(a) Answer: Gametes T and t from each parent; offspring genotypes TT, Tt, Tt, tt.
(b) B1 1 TT : 2 Tt : 1 tt (accept 1:2:1) ft from (a)
(b) Answer: 1 TT : 2 Tt : 1 tt
(c) B1 3 tall : 1 short ft from (a)
(c) B1 75% cao
(c) Answer: Phenotype ratio 3 tall : 1 short; 75% of offspring expected to be tall.

Question 5

(a) M1 correct gametes shown for each parent: F and f
(a) A1 correct offspring genotypes shown: FF, Ff, Ff, ff
(a) B1 ff correctly identified as the only genotype with cystic fibrosis
(a) Answer: Offspring genotypes FF, Ff, Ff, ff; only ff has cystic fibrosis.
(b) B1 25% (1 in 4) cao, ft from (a)
(b) Answer: 25% (1 in 4)
(c) B1 both parents are heterozygous carriers; the recessive allele (f) is masked by the dominant allele (F) in the parents, but a child can inherit the recessive allele from both parents (oe)
(c) Answer: Both parents are unaffected carriers (Ff); the recessive allele is hidden by the dominant allele in them, but a child can inherit a recessive allele from each parent.

Question 6

(a) M1 correct gametes shown: X and X from mother, X and Y from father
(a) A1 correct offspring genotypes shown (XX, XX, XY, XY) with probability of male = probability of female = 50% (1:1)
(a) Answer: Offspring are XX, XX, XY, XY; probability of a boy = 50%, probability of a girl = 50%.
(b) B1 the mother can only produce gametes containing an X chromosome
(b) B1 the father produces gametes containing either an X or a Y chromosome, so whichever gamete fertilises the egg determines whether the child is XX (female) or XY (male)
(b) Answer: The mother's eggs always carry an X chromosome, but the father's sperm carry either an X or a Y chromosome, so it is the father's gamete that determines the child's sex.

Question 7

(a) M1 correct offspring genotypes shown: Dd, Dd, dd, dd
(a) A1 50% cao
(a) Answer: 50%
(b) B1 natural selection removes alleles that reduce an organism's chance of survival or reproduction (i.e. that are selected against)
(b) B1 because polydactyly does not significantly affect survival or reproduction, people with the allele are just as likely to reproduce and pass it on, so it is not selected against/eliminated from the population
(b) Answer: Natural selection only removes alleles that reduce survival or reproductive success; as polydactyly does not do this, carriers reproduce as successfully as anyone else, so the allele is not eliminated from the population.

Question 8

Level 1 (1-2): Simple statements are made for or against embryonic screening, with little reasoning and no real balance.
Level 2 (3-4): Points are made on both sides of the argument, with some reasoning or justification, though the discussion may be one-sided or lack a clear conclusion.
Level 3 (5-6): A balanced, well-reasoned evaluation is given, covering benefits and drawbacks with scientific and ethical reasoning, leading to a justified conclusion.
Indicative content:
- Benefit: prevents parents passing on a serious genetic disorder to their child, reducing the child's future suffering
- Benefit: can reduce long-term emotional and financial strain on families and on healthcare services (e.g. NHS) from caring for a child with a severe genetic condition
- Benefit: gives parents who are known carriers the chance to have genetically related, unaffected children
- Drawback: the procedure (embryo biopsy, IVF) carries some risk and is expensive, and is not available to everyone
- Drawback: raises the ethical concern of 'designer babies' - screening could in principle be extended beyond serious disorders to select other, non-medical characteristics
- Drawback: some people view discarding embryos with a genetic disorder as ethically unacceptable, or as suggesting that people currently living with that disorder are less valued
- A balanced conclusion should weigh the reduction in suffering/cost against the ethical concerns and limited access, without necessarily coming down firmly on one side

Question 9

(a) B1 genetic causes (differences in the alleles/genes an individual inherits)
(a) B1 environmental causes (differences in the conditions an individual lives/develops in)
(a) Answer: Genetic causes (differences in inherited alleles) and environmental causes (differences in living conditions).
(b) B1 e.g. an accent, a scar, a suntan, or a plant's height/leaf colour due to differences in light or nutrient availability (any suitable environmental example)
(b) Answer: For example, a scar, a suntan, or a plant growing taller because it received more light or fertiliser.

Question 10

(a) B1 a change in the DNA base sequence of a gene (oe)
(a) Answer: A change in the DNA base sequence of a gene.
(b) B1 most mutations occur in regions of DNA that do not code for a characteristic, or do not change the protein produced enough to affect its function (oe)
(b) B1 so the protein made still works normally/the same characteristic is still produced, meaning the phenotype is unaffected
(b) Answer: Most mutations do not significantly change the protein produced, so the protein still functions normally and the phenotype is not noticeably affected.

Question 11

(a) B1 lay a tape measure/grid across the habitat and use a random number generator (or random number table) to choose coordinates for quadrat placement
(a) B1 place the quadrat at each random coordinate and count/record the snails found inside it, repeating this for a suitable number of quadrats across the habitat
(a) Answer: Use random numbers to choose coordinates along a grid laid across the habitat, then place the quadrat at each coordinate and record the snails found, repeating for several quadrats.
(b) B1 e.g. time of day, time of year/season, weather conditions, or sampling effort (number of quadrats/area sampled) - any one sensible controlled variable
(b) Answer: For example, sampling at the same time of year and in similar weather conditions in both habitats.
(c) M1 (32 / 40) x 100
(c) A1 80% cao
(c) Answer: 80%
(d) B1 75% cao [(45/60) x 100]
(d) Answer: 75%
(e) B1 predators (e.g. thrushes) hunt snails by sight, and snails that are less well camouflaged in their habitat are more likely to be seen and eaten
(e) B1 in the hedgerow, the patchy light and shade mean banded shells are better camouflaged, while in the uniform short grassland, plain (unbanded) shells are better camouflaged
(e) B1 in the grassland the banded/striped shell gives better camouflage against the streaky grass background, so banded snails are less likely to be predated by birds, more likely to survive and reproduce, and pass on the allele for banding; this selection pressure explains the higher percentage of banded snails in the grassland (80%) than in the hedgerow (75%) (oe reasoning credited)
(e) Answer: The shell pattern that is best camouflaged against each background is less likely to be spotted and eaten by predators, so it survives and reproduces more successfully, making that pattern more common in that habitat over time.
(f) B1 a larger sample size reduces the effect of chance/random variation on the results (e.g. from sampling a few snails in one patch by chance)
(f) B1 so the calculated percentages are more likely to be representative of the true population in each habitat, making the comparison between habitats more valid
(f) Answer: A bigger sample reduces the effect of chance, so the percentages calculated are more likely to reflect the true population, making the comparison more valid.
(g) B1 e.g. the populations may be isolated with little migration between them (so allele frequencies can drift apart by chance), or one population may have started with a different allele frequency by chance (founder effect) - any sensible alternative explanation
(g) Answer: For example, the two populations may be isolated from each other, so random chance (genetic drift) or the starting allele frequencies could differ, independent of camouflage.
(h) B1 the process by which organisms with characteristics (e.g. shell pattern) best suited to their environment are more likely to survive and reproduce, passing those characteristics on to their offspring (oe, in context)
(h) Answer: The process by which the best-camouflaged (best-adapted) snails are more likely to survive, reproduce and pass on their shell-pattern allele.
(i) B1 other factors could also affect the observed percentages, e.g. differences in predator density between the two sites, or migration of snails between habitats, which are not measured in this investigation (oe)
(i) B1 so a difference in banding percentage alone does not prove that camouflage/predation (natural selection) is the cause, without ruling out these other factors
(i) Answer: Other factors, such as differences in predator numbers or migration between sites, could also explain the pattern; the data alone cannot rule these out as the true cause.

Question 12

B1 individuals in a population show variation, e.g. some are better camouflaged than others, due to differences in their alleles
B1 organisms compete for limited resources (e.g. food, space, mates); better camouflaged individuals are less likely to be caught by predators
B1 individuals with characteristics best suited to the environment are more likely to survive and reproduce (survival of the fittest)
B1 the alleles for the advantageous characteristic are passed on to offspring, so over many generations the proportion of the population with that characteristic increases
Answer: Indicative content - award any 4 of the marking points listed, max 4 marks.

Question 13

(a) M1 (250 / 5 000 000) x 100
(a) A1 0.005% cao
(a) Answer: 0.005%
(b) B1 the antibiotic killed the non-resistant bacteria, as they had no mutation giving resistance
(b) B1 the resistant bacteria survived the antibiotic treatment because their mutation protected them from its effects
(b) B1 the surviving resistant bacteria then reproduced (without competition from the non-resistant bacteria), so nearly all of the remaining/new population was descended from resistant bacteria
(b) Answer: The antibiotic killed the non-resistant bacteria; the resistant bacteria survived and then reproduced, so almost all of the surviving population came from resistant ancestors.

Question 14

(a) B1 from the hard parts of an organism (e.g. bones, shells, teeth) that do not decay easily, and which may be gradually replaced by minerals as they are buried
(a) B1 from organisms preserved in places where decay is prevented, e.g. in amber, tar pits, peat bogs, or ice (any one suitable example)
(a) Answer: Fossils can form from hard body parts replaced by minerals over time, or from whole organisms preserved in places where decay is prevented, such as amber or ice.
(b) B1 many early forms of life were soft-bodied, which decayed completely and left few traces/fossils
(b) B1 many fossils that did form have since been destroyed by geological activity (e.g. rock movement, erosion) before they could be discovered
(b) Answer: Soft-bodied organisms rarely fossilise, and many fossils that did form have since been destroyed by geological processes, so the record is incomplete.

Question 15

(a) B1 Archaea, Bacteria and Eukarya (all three needed)
(a) Answer: Archaea, Bacteria and Eukarya.
(b) B1 advances in microscopy and biochemical/chemical analysis allowed the internal structures and molecules (e.g. DNA/RNA sequences) of organisms to be studied in far more detail than before
(b) B1 this evidence showed that organisms previously classed together in the kingdom Prokaryotae/Monera (e.g. certain bacteria-like organisms) were, at a molecular level, as different from each other as they were from eukaryotic organisms
(b) B1 this meant the five-kingdom system did not accurately reflect how closely related organisms actually were, so the three-domain system was proposed, splitting the former prokaryotes into Archaea and Bacteria
(b) Answer: Improved biochemical and DNA/RNA evidence showed that some organisms grouped together as prokaryotes were less closely related than thought, so classification was revised into three domains to better reflect evolutionary relationships.

Question 16

(a) B1 select existing wheat plants that show the most disease resistance
(a) B1 breed these selected plants together
(a) B1 from the offspring, select those showing the greatest disease resistance and breed these together
(a) B1 repeat this process over many generations, so that disease resistance gradually increases in the population
(a) Answer: Select the most disease-resistant plants, breed them together, then select and breed the most resistant offspring; repeat over many generations to increase resistance.
(b) B1 e.g. increases crop yield/reduces crop losses caused by disease, reducing costs for the farmer (or improves food security)
(b) Answer: It reduces crop losses to disease, increasing yield and reducing costs for the farmer.
(c) B1 reduces the gene pool/genetic variation in the wheat population, which can increase the risk of a different disease or environmental change affecting the whole crop (inbreeding depression) (oe)
(c) Answer: It reduces genetic variation in the population, making the whole crop more vulnerable to a new disease or environmental change.

Question 17

B1 the gene for human insulin is identified and isolated (cut) from human DNA
B1 a restriction enzyme is used to cut out the insulin gene, leaving 'sticky ends'
B1 a bacterial plasmid (vector) is cut open using the same restriction enzyme, producing complementary sticky ends
B1 the enzyme ligase is used to join (splice) the insulin gene into the plasmid vector
B1 the modified plasmid is inserted into a bacterial cell (transformation)
B1 the bacterium is grown/cultured, and it expresses the human gene to produce human insulin, which can then be collected and purified
Answer: Indicative content - award any 6 of the marking points listed, max 6 marks.

Question 18

(a) B1 fewer crops are lost/damaged by pests, increasing crop yield
(a) B1 less pesticide/insecticide needs to be sprayed on the crop, reducing costs and environmental impact
(a) Answer: GM pest-resistant crops lose less yield to pests and need less pesticide spraying, reducing costs and environmental impact.
(b) B1 the introduced gene could be transferred to wild plant species (e.g. via pollen), possibly creating herbicide-resistant or otherwise problematic 'superweeds'
(b) B1 some people have concerns about the long-term effects of eating GM crops on human health, or about the reduction in biodiversity from growing large areas of a single GM variety, or the cost of GM seed to farmers (any one further concern)
(b) Answer: For example, the modified gene could spread to wild plants, or there may be concerns about long-term health effects, reduced biodiversity, or seed cost.

Question 19

(a) M1 all offspring genotype Rr
(a) A1 all offspring phenotype pink
(a) B1 codominance means both alleles (R and r) are expressed in the phenotype of the heterozygote, giving an intermediate/blended phenotype (pink), rather than one allele being completely dominant over/masking the other
(a) Answer: All offspring are Rr and pink; codominance means both alleles are expressed together, producing a blended phenotype rather than one allele masking the other.
(b) M1 correct offspring genotypes shown: RR, Rr, Rr, rr
(b) A1 genotype ratio 1 RR : 2 Rr : 1 rr
(b) B1 phenotype ratio 1 red : 2 pink : 1 white
(b) Answer: Genotype ratio 1 RR : 2 Rr : 1 rr; phenotype ratio 1 red : 2 pink : 1 white.

Question 20

(a) M1 expected ratio 3 tall : 1 short applied to 200: (3/4) x 200 and (1/4) x 200
(a) A1 150 tall and 50 short cao
(a) Answer: 150 tall plants and 50 short plants.
(b) B1 which gamete fertilises which is a matter of chance, so with a limited/finite sample the actual ratio produced will not exactly match the theoretical ratio (oe)
(b) B1 a larger sample size would be expected to give a ratio closer to the theoretical 3:1 ratio, as random variation would have less effect on the overall proportions
(b) Answer: Which allele combination occurs is down to chance, so a finite sample rarely matches the exact theoretical ratio; a larger sample would be expected to be closer to 3:1.

Inheritance, Variation and Evolution - Worksheets, Questions and Revision