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Form 5 Biology Practice Paper: Genetics, Molecular Biology and Ecology
Form 5 Biology – 12 questions – 100 marks – 75 minutes
Form 5 Biology practice paper: 12 questions, 100 marks, 75 minutes, covering genetics, DNA, PCR, energy flow, sampling and succession, with full answers.
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Paper details Form 5 Biology practice paper with 12 structured questions on genetics, molecular biology and ecology; includes a full marking scheme and two printable PDFs.
Section A: Genetics (Q1-Q4)
Answer all parts. Show all working, steps and units in calculations.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | In garden peas, tall stem (T) is dominant over dwarf stem (t). A monohybrid cross is made between a homozygous tall plant and a homozygous dwarf plant to give the F1 generation; the F1 plants are then self-pollinated and produce 640 F2 plants. (a) State the genotype and phenotype of the F1 generation. (b) Use a genetic diagram to show the genotypic ratio of the F2 generation. (c) Calculate the expected numbers of tall and dwarf plants in the F2 generation. (d) If an F1 plant is crossed with a homozygous dwarf plant, what phenotypic ratio is expected? |
(a) The F1 genotype is Tt and all F1 plants are tall, because T is dominant. (b) Tt x Tt gives gametes T and t; the offspring genotypes are TT : Tt : tt = 1 : 2 : 1, giving 3 tall : 1 dwarf. (c) Tall = 640 x 3 / 4 = 480 plants; dwarf = 640 x 1 / 4 = 160 plants. (d) Tt x tt gives Tt : tt = 1 : 1, so tall : dwarf = 1 : 1. This test cross shows the F1 genotype. Marks: correct gametes, genotypic ratio, the two numbers, test-cross ratio. |
| 2 | In peas, round seed (R) is dominant over wrinkled seed (r) and yellow cotyledon (Y) is dominant over green cotyledon (y); the two genes are on different chromosomes. Two dihybrid plants (RrYy) are crossed and 800 offspring are obtained. (a) State the types and ratio of gametes produced by a dihybrid plant. (b) State the phenotypic ratio of this cross. (c) Calculate the expected number of each phenotype. (d) One wrinkled yellow offspring appears. State its possible genotypes and explain how a test cross identifies the genotype. |
(a) The gametes are RY, Ry, rY and ry in the ratio 1 : 1 : 1 : 1, because the genes assort independently. (b) The ratio is 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. (c) Round yellow = 800 x 9 / 16 = 450; round green = 150; wrinkled yellow = 150; wrinkled green = 800 x 1 / 16 = 50. (d) The genotype is rrYY or rrYy. Cross it with a homozygous wrinkled green plant (rryy): if all offspring are wrinkled yellow the parent is rrYY; if wrinkled green offspring appear the parent is rrYy. Marks: four gametes, 9:3:3:1 ratio, four numbers, test-cross conclusion. |
| 3 | In the human ABO blood group system, alleles IA and IB show codominance and allele i is recessive. A group A woman (IA i) marries a group B man (IB i) and they have four children. (a) List the gametes of the parents and the possible genotypes of the children. (b) State the possible blood groups of the children and their ratio. (c) Among 400 couples of the same cross, each with one child, calculate the expected number of children of each blood group. (d) The couple already has a group O child. What is the probability that the next child is group AB? |
(a) The mother produces IA and i gametes and the father produces IB and i gametes; fertilisation gives IA IB, IA i, IB i and ii. (b) The blood groups are AB, A, B and O in the ratio 1 : 1 : 1 : 1, showing codominance and recessive inheritance. (c) 400 / 4 = 100 children of each group. (d) Each fertilisation is an independent event, so the probability is still 1 / 4 = 0.25, that is 25%. Marks: gametes and genotypes, blood groups and ratio, calculation, independence of events. |
| 4 | Red-green colour blindness is a sex-linked condition caused by a recessive allele on the X chromosome. A woman who carries the allele but has normal vision (XB Xb) marries a man with normal vision (XB Y). (a) Write the genotypes of the parents and the possible genotypes of their children. (b) State the proportion of colour-blind sons and of colour-blind daughters. (c) If the couple has 4 children, find the probability that all 4 have normal vision. (d) Among 400 families in the same situation, how many are expected to have all 4 children with normal vision? |
(a) The mother is XB Xb and the father is XB Y. Sons are XB Y or Xb Y; daughters are XB XB or XB Xb. (b) Half of the sons (1/2) are colour blind; all daughters have normal vision but half of them are carriers. (c) The probability for one child is 3/4, so for 4 children it is (3/4)^4 = 81/256 = 0.316. (d) 400 x 0.316 = 126.6, so about 127 families. Marks: genotypes, sons and daughters, probability, number of families. |
Section B: Molecular Genetics and Biotechnology (Q5-Q8)
Write your answers in full sentences; explanations must show the cause and effect.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A polypeptide consists of 150 amino acids; each amino acid is coded by one codon and carried by one tRNA. (a) Find the minimum number of bases in the coding region of the mRNA for this polypeptide. (b) Find the number of bases needed on the template strand and the number of base pairs in the whole gene. (c) Explain the effect on the polypeptide if the 60th codon changes from UAC to UAA. (d) Why is the actual length of a gene usually greater than the value calculated in (b)? |
(a) Three bases code for one amino acid, so the coding region has at least 150 x 3 = 450 bases. (b) The template strand also needs 450 bases; the double-stranded gene has 450 base pairs, that is 900 bases. (c) UAA is a stop codon, so translation stops before the 60th amino acid and the polypeptide has only 59 amino acids; it usually loses its function. (d) A gene also contains a promoter, a terminator, introns and non-coding control sequences; introns are removed by splicing, so the gene is longer than the coding region. Marks: calculation, 450 and 900, stop codon effect, non-coding sequences. |
| 6 | The polymerase chain reaction is used to amplify a specific DNA fragment. The reaction starts with a single DNA molecule and runs for 20 cycles. (a) State the three main steps of each cycle with their temperature ranges. (b) Calculate the number of DNA molecules after 20 cycles. (c) Explain why a heat-stable DNA polymerase is needed. (d) If only 1 000 molecules are needed, what is the minimum number of complete cycles? |
(a) Denaturation at about 95 degrees separates the strands; annealing at about 55 to 65 degrees lets primers bind; extension at about 72 degrees synthesises the new strand. (b) The number doubles each cycle, so after 20 cycles there are 2^20 = 1,048,576 molecules. (c) The high temperature of denaturation would denature an ordinary enzyme, while a heat-stable polymerase such as Taq remains active. (d) 2^10 = 1024, so at least 10 complete cycles are needed. Marks: three steps and temperatures, number of molecules, heat stability, cycle number. |
| 7 | A circular plasmid is 4,000 bp long. Digestion with the restriction enzyme EcoRI (recognition site GAATTC) gives two fragments of 1,800 bp and 2,200 bp on a gel. (a) State the number of EcoRI recognition sites on the plasmid and explain why two fragments appear. (b) Add the two fragment lengths and check whether the total equals the plasmid length. (c) State which fragment moves faster in gel electrophoresis and explain why. (d) State two requirements for inserting one fragment into another plasmid to make recombinant DNA. |
(a) There are two EcoRI sites; cutting a circular plasmid twice produces two fragments. (b) 1,800 + 2,200 = 4,000 bp, which equals the plasmid length. (c) The 1,800 bp fragment moves faster because it is shorter and is retarded less by the gel network. (d) The vector must be cut with the same restriction enzyme to give complementary ends, and DNA ligase is needed to join the fragment to the vector; the recombinant plasmid also needs an origin of replication and a selectable marker. Marks: two sites, length check, mobility reason, two requirements. |
| 8 | The producers of a grassland fix 48,000 kJ m-2 per year. The efficiency of energy transfer between adjacent trophic levels is 10%. (a) Calculate the energy available to the primary, secondary and tertiary consumers. (b) State three fates of the remaining 90% of the energy. (c) Explain why food chains rarely have more than four or five trophic levels. (d) If the grassland is converted to farmland, what happens to energy transfer and to the number of trophic levels? |
(a) Primary consumers = 48,000 x 10% = 4,800; secondary consumers = 480; tertiary consumers = 48 kJ m-2 per year. (b) Energy is lost as heat from respiration, lost in excretory products and dead remains, and passes to decomposers through uneaten material. (c) About 90% is lost at each transfer, so the energy available falls quickly and higher trophic levels cannot support a viable population. (d) A field of one crop concentrates energy in one trophic level and shortens the food chain, but biodiversity and stability are lower. Marks: three values, three fates, energy loss explanation, farmland comparison. |
Section C: Ecology and Applications (Q9-Q12)
Use standard symbols for genotypes; in experimental design identify the independent and controlled variables.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A mark-recapture study estimates the number of tilapia in a pond: 200 fish are caught, marked and released; one week later 250 fish are caught and 40 of them carry a mark. (a) Write the estimating formula and calculate the population size. (b) State three assumptions of the method. (c) If marks are easily lost, is the estimate too high or too low? Explain. (d) What happens to the estimate if the second sample is taken only in shallow water near the bank? |
(a) Population size = (number marked x number in second sample) / number marked in second sample = (200 x 250) / 40 = 1,250 fish. (b) The population is closed, marking does not affect survival or the chance of recapture, and marked individuals mix evenly with the population. (c) Lost marks reduce the number of marked fish in the second sample, so the denominator is smaller and the estimate is too high. (d) If fish are not evenly distributed, the second sample does not represent the whole population, so the estimate is biased and usually too low. Marks: formula and value, three assumptions, high estimate, sampling bias. |
| 10 | A meadow of 2,500 m2 is sampled with quadrats. Fifty random 0.5 m2 quadrats contain on average 6.4 plants each. (a) Calculate the total sampled area and the population density in plants per m2. (b) Estimate the population size of the whole meadow. (c) Explain why random sampling is important and state one way to ensure it. (d) What bias arises if quadrats are placed only along the path? |
(a) Sampled area = 50 x 0.5 = 25 m2; density = 6.4 / 0.5 = 12.8 plants per m2. (b) Population = 12.8 x 2,500 = 32,000 plants. (c) Random sampling avoids bias from personal choice; for example, divide the meadow into a grid and use random numbers to choose quadrat coordinates. (d) The path is disturbed, so its density differs from the rest of the meadow and the estimate is systematically biased. Marks: area and density, total number, random method, bias. |
| 11 | In a population at Hardy-Weinberg equilibrium, a recessive allele a has a frequency of 0.3 and the population has 5,000 individuals. (a) Find the frequency of the dominant allele A. (b) Calculate the frequencies of the three genotypes. (c) Estimate the number of individuals showing the recessive trait and the number of carriers. (d) Explain how inbreeding would change these frequencies. |
(a) p = 1 – q = 1 – 0.3 = 0.7. (b) AA = p2 = 0.49; Aa = 2pq = 2 x 0.7 x 0.3 = 0.42; aa = q2 = 0.09, and the total is 1.00. (c) Recessive trait = 5,000 x 0.09 = 450; carriers = 5,000 x 0.42 = 2,100. (d) Inbreeding does not change allele frequencies but increases homozygotes and reduces heterozygotes, so the recessive trait appears more often. Marks: p value, three frequencies, numbers, inbreeding effect. |
| 12 | An abandoned field undergoes secondary succession for forty years. (a) Arrange the stages in order: shrubs, pioneer herbs, tree woodland, mosses and lichens. (b) Describe the trends in species diversity, biomass and soil organic matter during succession. (c) State two features that allow pioneer species to survive on bare ground. (d) Design a three-year survey comparing species richness in an early and a late stage plot; state the independent variable, the dependent variable and the sampling method. |
(a) Mosses and lichens, then pioneer herbs, then shrubs, then tree woodland, a clear sequence of succession. (b) Species diversity rises, total biomass increases, and soil organic matter and water content increase until the climax community becomes stable. (c) Pioneers tolerate strong light, drought and poor soil and reproduce and disperse quickly; lichens also release acids that break down rock to form soil. (d) The independent variable is the successional stage (early and late plots); the dependent variable is species richness, the number of species in a plot; place ten random 1 m2 quadrats in each plot, count the species in each quadrat and calculate the mean, surveying once each season for three years. Marks: order, three trends, pioneer features, survey design. |
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