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Form 6 Biology Practice Paper: Genetics, DNA and Evolution
Form 6 Biology – 12 questions – 100 marks – 75 minutes
Form 6 Biology paper: 12 questions, 100 marks, 75 minutes on crosses, chi-square, Hardy-Weinberg, base composition, transcription, translation and mark-recapture.
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Paper details Form 6 Biology practice paper with 12 structured questions on monohybrid and dihybrid crosses, test cross, chi-square, Hardy-Weinberg equilibrium, DNA and mark-recapture.
Section A: Mendelian Inheritance and the Chi-square Test (Q1-Q4)
Answer all questions; show the expression, substitution and unit for every calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Stem height in pea plants is controlled by one pair of alleles, tall (T) dominant to short (t). Two tall plants, both of genotype Tt, are crossed and produce 400 offspring, of which 300 are tall and 100 are short. This is a monohybrid cross. (a) State the expected genotypic ratio of the offspring. (b) Using 400 offspring as the base, calculate the expected numbers of tall and short plants. (c) Among all tall offspring, what fraction is expected to be homozygous tall (TT)? (d) If only 40 offspring were obtained, must exactly 30 of them be tall? Explain. |
(a) TT : Tt : tt = 1 : 2 : 1, that is 1/4 TT, 2/4 Tt and 1/4 tt. (b) Tall = 400 x 3/4 = 300; short = 400 x 1/4 = 100. (c) Among tall offspring TT : Tt = 1 : 2, so the fraction of homozygous tall is 0.333 (1/3). (d) No. Each offspring is an independent random combination, so the number of tall plants follows a binomial distribution. With 40 offspring the expected number is 30, but 26 or 34 can occur. Only with a large sample does the observed ratio approach 3 : 1, and a chi-square test is needed to judge significance. Marking notes: (a) 2 marks; (b) 2 marks; (c) 2 marks; (d) conclusion 1 mark, explanation 2 marks. |
| 2 | Seed shape (round R dominant to wrinkled r) and cotyledon colour (yellow Y dominant to green y) in peas are controlled by two pairs of alleles on different chromosomes. Two plants, both RrYy, are crossed and produce 1600 seeds. This is a dihybrid cross. (a) State the expected phenotypic ratio in the second filial generation. (b) Calculate the expected number of seeds of each of the four phenotypes. (c) Calculate the proportion and the number of recombinant types (round wrinkled plus wrinkled green). (d) Why is the location of the two gene pairs on different chromosomes necessary for a 9 : 3 : 3 : 1 ratio? |
(a) Round yellow : round wrinkled : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1. (b) Round yellow = 1600 x 9/16 = 900; round wrinkled = 300; wrinkled yellow = 300; wrinkled green = 100. (c) Recombinants = 3/16 + 3/16 = 0.375, that is 1600 x 0.375 = 600 seeds. (d) Because the two pairs lie on different chromosomes, they assort independently at meiosis, so the four gametes RY, Ry, rY and ry are produced in equal numbers (1/4 each). With linkage the gamete ratio departs from 1 : 1 : 1 : 1 and the 9 : 3 : 3 : 1 ratio breaks down. Marking notes: (a) 2 marks; (b) 3 marks; (c) 2 marks; (d) 2 marks. |
| 3 | A black male mouse is crossed with a brown female. Coat colour is controlled by one pair of alleles, black (B) dominant to brown (b). The cross produces 160 offspring, 80 black and 80 brown. (a) Deduce the genotypes of the two parents and explain your reasoning. (b) What is this breeding design called in genetics, and what is its purpose? (c) If the male had in fact been BB, what phenotypes and numbers would the offspring show? (d) Give the expected number of offspring of each genotype from the cross in (a). |
(a) The female must be bb because brown is the homozygous recessive phenotype. Brown offspring are bb, so each received one b from the father; the male is black, so he must be Bb. Check: Bb x bb gives 1 Bb : 1 bb, that is half black and half brown, which is 80 and 80 out of 160. This is a test cross. (b) A test cross: the individual of unknown genotype is crossed with a homozygous recessive and the phenotypic ratio of the offspring reveals the unknown genotype. (c) All offspring of BB x bb are Bb and therefore black: 160 black and none brown. (d) Bb 80; bb 80. Marking notes: (a) each genotype 1 mark, reasoning 2 marks; (b) 2 marks; (c) 2 marks; (d) 1 mark. |
| 4 | A horticulturist crosses two pink-flowered plants. The genetic hypothesis predicts a red : pink : white ratio of 1 : 2 : 1. In 720 offspring the observed numbers are 190 red, 355 pink and 175 white. (a) Calculate the expected number for each phenotype under the 1 : 2 : 1 hypothesis. (b) Calculate the chi-square value to two decimal places. (c) How many degrees of freedom are there? The critical value at the 5% level is 5.99; should the hypothesis be rejected? (d) If the hypothesis is not rejected, does that prove it is correct? Explain. |
(a) Red = 720 / 4 = 180; pink = 720 x 2/4 = 360; white = 180. (b) chi-square = sum of (O – E)^2 / E = 100/180 + 25/360 + 25/180 = 0.76. (c) Degrees of freedom = number of classes – 1 = 2. Since 0.76 is below the critical value 5.99, the hypothesis is not rejected. (d) Not rejecting is not proof. It only means the data do not differ significantly from 1 : 2 : 1. A larger sample might still reveal a significant deviation, and other hypotheses could also fit the data, so the result merely shows the data are compatible with the hypothesis. Marking notes: (a) 2 marks; (b) 3 marks; (c) 2 marks; (d) 1 mark. |
Section B: Population Genetics, DNA and Protein Synthesis (Q5-Q8)
Data: for 2 degrees of freedom the critical chi-square value at the 5% level is 5.99; under Hardy-Weinberg equilibrium the recessive homozygote frequency is q squared.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A recessive genetic disease affects 1 person in 10,000 in a large population that is in Hardy-Weinberg equilibrium (random mating, no mutation, no migration, no selection). (a) Find the recessive allele frequency q and the dominant allele frequency p. (b) Find the carrier (heterozygote) frequency as a percentage. (c) Estimate the number of carriers in a population of 50,000. (d) Why are there far more carriers than patients, and what does this mean for controlling the disease? |
(a) The patient genotype frequency is q squared = 1/10,000, so q = 0.0100 and p = 1 – q = 0.99. (b) Carrier frequency = 2pq = 2 x 0.99 x 0.0100 = 0.0198, about 1.98%. (c) Carriers = 50,000 x 0.0198 = 990 people. (d) A carrier has only one copy of the allele, and the recessive allele is not expressed, so selection does not remove it. The carrier frequency depends on 2pq rather than q squared, and when p is close to 1 the value of 2pq is far larger than q squared. Screening of carriers and genetic counselling are therefore needed, because testing patients alone cannot identify carriers or estimate the risk to the next generation. Marking notes: (a) 2 marks; (b) 2 marks; (c) 2 marks; (d) 3 marks. |
| 6 | A double-stranded DNA molecule is 3000 base pairs long. Chemical analysis shows that adenine (A) makes up 28% of all the bases. (a) Find the percentage of thymine (T), guanine (G) and cytosine (C) in the whole molecule. (b) Calculate the actual number of G bases in this molecule. (c) If adenine makes up 20% of one strand, what percentage of the other strand is adenine? (d) From base composition alone, can you tell whether this DNA comes from a human or a bacterium? Explain. |
(a) In double-stranded DNA A = T, so T is also 28%. Then G + C = 100% – 56% = 44%, and because G = C, both are 22%. (b) Total bases = 3000 x 2 = 6000, so G = 6000 x 0.22 = 1320 bases. (c) One strand has 3000 bases and 20% of them are A, that is 600 bases. The whole molecule has 1680 A bases, so the other strand has 1080 A bases, which is 36% of that strand. (d) No. The base composition has no species specificity and different organisms can have similar values. Base composition is a measure of the whole genome; identifying the source requires the actual base sequence, and the genome size and sequence of a human and a bacterium differ enormously. Marking notes: (a) 3 marks; (b) 2 marks; (c) 2 marks; (d) 1 mark. |
| 7 | Meselson and Stahl grew E. coli for many generations in a medium containing heavy nitrogen (15N), then transferred the bacteria to a medium containing only 14N. The bacteria divide every 20 minutes. (a) At the first, second and third generation, how many DNA molecules are there and how many of them are hybrid (15N/14N)? (b) What percentage of the DNA molecules are hybrid at the third generation? (c) Which model of replication do the results support? Explain using the first generation. (d) If DNA replicated conservatively, what would be seen at the first generation? |
(a) First generation: 2 molecules, all 2 hybrid. Second generation: 4 molecules, 2 hybrid. Third generation: 8 molecules, 2 hybrid. The number of hybrid molecules stays at 2 while the total doubles. (b) 25% (2 out of 8). (c) Semiconservative replication. At the first generation every molecule is hybrid, showing that each daughter molecule keeps one parental (15N) strand and has one newly made (14N) strand; no fully heavy or fully light molecule appears. (d) With conservative replication the parental double helix would stay intact and a separate all-new molecule would be made, so the first generation would show one fully heavy (15N) molecule and one fully light (14N) molecule with no hybrid, which contradicts the observation. Marking notes: (a) 3 marks; (b) 1 mark; (c) 3 marks; (d) 2 marks. |
| 8 | An mRNA molecule is 300 nucleotides long, excluding the poly-A tail. It carries one stop codon at the 3′ end and the rest is coding sequence. (a) How many codons does this mRNA contain? (b) How many amino acids can be translated from it? (c) If one nucleotide is inserted in the middle of the coding region, what usually happens to the product, and why? (d) If one base on the template strand changes from T to C, how does the mRNA change, and does this always change the amino acid? |
(a) Number of codons = 300 / 3 = 100, including the stop codon. (b) The stop codon does not code for an amino acid, so 100 – 1 = 99 amino acids are translated. (c) A frameshift results. After the insertion every codon downstream is read in a new reading frame, so the amino acid sequence changes drastically and the protein is usually non-functional or terminated early. (d) T on the template strand pairs with A, and C pairs with G, so the mRNA base changes from A to G at that position. Not necessarily: the genetic code is degenerate, several codons may specify the same amino acid, and a change in the third base is often a silent mutation. A missense or nonsense mutation can also occur. Marking notes: (a) 1 mark; (b) 1 mark; (c) 4 marks; (d) 3 marks. |
Section C: Measurement, Blood Groups, Evolution and Ecology (Q9-Q12)
Give numerical answers to three significant figures and express probabilities as both a fraction and a decimal.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A student observes a transverse section of a leaf with a light microscope. A cell measures 45 divisions on the eyepiece graticule, and at that total magnification each division represents 20 micrometres. (a) Find the actual length of the cell in micrometres. (b) The cell measures 45 mm on a micrograph. Find the magnification of the photograph. (c) The student changes the objective from 10x to 40x while the eyepiece stays at 10x. Find the total magnification before and after the change and state how the diameter of the field of view changes. (d) Calculate the factor by which the area of the field of view is reduced. |
(a) Actual length = 45 x 20 = 900 micrometres (0.90 mm). (b) Magnification = image length / actual length = 45 mm / 0.90 mm = 50 times. (c) Before: 10 x 10 = 100 times. After: 10 x 40 = 400 times. The magnification is 4 times greater, so the diameter of the field of view falls to one quarter. (d) Area varies with the square of the diameter, so the area is reduced 16 times, that is to one 16th of the original. Marking notes: (a) 2 marks; (b) 2 marks; (c) 3 marks; (d) 1 mark. |
| 10 | Human ABO blood groups are controlled by three alleles, IA, IB and i. IA and IB are codominant and i is recessive. A man of group AB marries a woman of group O. (a) Give the genotypes of the parents and of the possible offspring. (b) Give the ratio of the four blood groups (A, B, AB, O) among the offspring. (c) They plan three children. Find the probability that at least one child is group B. (d) If the man were group A with genotype IAi, what ratio of blood groups would his children have? |
(a) The man is IAIB and the woman is ii. The offspring can only be IAi (group A) or IBi (group B); IAIA, IBIB, IAIB and ii are all impossible. (b) A : B : AB : O = 1 : 1 : 0 : 0, that is 50% A and 50% B. (c) The probability that one child is group B is 0.5, so the probability that none of three children is group B is (0.5)^3 = 0.125. Hence the probability that at least one is group B is 1 – 0.125 = 0.875, or 7/8. (d) IAi x ii gives IAi and ii in equal numbers: 50% group A and 50% group O, with no group B or AB child. Marking notes: (a) 2 marks; (b) 2 marks; (c) 2 marks; (d) 2 marks. |
| 11 | Coat colour in a lizard population on an island is controlled by one pair of alleles, dark (D) dominant to light (d). In 1970 the allele frequency of D was 0.35; in 2010 it had risen to 0.62. (a) Find the frequency of the light allele d in 1970 and in 2010. (b) Calculate the percentage change in the allele frequency of D over these 40 years (three significant figures). (c) If the population stays at 2,000 and is in Hardy-Weinberg equilibrium, find the number of light individuals in 2010. (d) Suggest one selection pressure that could explain the change and explain how it alters phenotype frequency. |
(a) In 1970 the frequency of d was 1 – 0.35 = 0.65; in 2010 it was 1 – 0.62 = 0.38. (b) Percentage change = (0.62 – 0.35) / 0.35 x 100% = 77.1%. (c) Light lizards are dd, with frequency q squared = 0.38 squared = 0.1444, so the number is 2,000 x 0.1444 = 288.8, about 289 individuals. (d) For example the rocks became darker after volcanic activity. Dark lizards are harder for birds to see on a dark background, so they survive and breed more successfully and leave more offspring, while light individuals are caught more often and the d allele is gradually removed. The frequency of the dark phenotype rises, which is directional selection. Marking notes: (a) 2 marks; (b) 2 marks; (c) 2 marks; (d) 2 marks. |
| 12 | An ecologist estimates a snail population by the mark-recapture method in a plot of 0.25 square kilometres. 120 snails are caught, marked and released; a week later 150 snails are caught and 18 of them carry a mark. (a) Estimate the total number of snails in the plot. (b) Calculate the population density in individuals per square kilometre. (c) Give two factors that would make this estimate too high. (d) If 30 of the 150 snails in the second sample had been marked, what would the estimate be, and what might that mean? |
(a) N = (number marked x number in second sample) / number marked in second sample = (120 x 150) / 18 = 1000 snails. (b) Density = 1000 / 0.25 = 4000 per square kilometre. (c) Anything that lowers the proportion of marked animals in the second sample raises the estimate: marks may make snails more visible to predators or affect movement so marked animals die or leave the plot, marked individuals may learn to avoid the traps, or many unmarked immigrants may enter the plot between the two samples. (d) N = (120 x 150) / 30 = 600 snails. The estimate falls, which may mean the population is really smaller, or that marks were lost, or that marked snails were easier to recapture (trap happy), making the proportion of marked animals too high and the estimate too low. Marking notes: (a) 2 marks; (b) 2 marks; (c) 2 marks; (d) 3 marks. |
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