Free Inheritance questions for A Level Biology (9700), part of Genetics & evolution. Each question has the answer and the marking points an examiner would look for.
Generate a full Inheritance worksheet →All Genetics & evolution questionsMeiosis produces genetic variation.
Explain how crossing over and independent assortment produce variation.
| Part | Answer and marking points | Marks |
|---|---|---|
| exchange between chromatids; random orientation of pairs crossing over at chiasmata in prophase I B1 exchanges alleles between non-sister chromatids B1 independent assortment: random orientation of homologous pairs at metaphase I B1 produces different combinations of alleles B1 | 4 |
In a plant, purple flowers (P) are dominant to white (p), and tall stems (T) are dominant to short (t). The genes are on different chromosomes. Two plants heterozygous for both genes are crossed and 206 offspring are counted.
| phenotype | purple tall | purple short | white tall | white short |
|---|---|---|---|---|
| observed | 109 | 32 | 44 | 21 |
State the expected ratio and calculate the expected numbers.
Calculate the chi-squared value.
The critical value at p = 0.05 with 3 degrees of freedom is 7.82. State what the result shows.
Explain how crossing over and independent assortment in meiosis produce genetic variation.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 9 : 3 : 3 : 1; 116, 38.6, 38.6, 12.9 9 : 3 : 3 : 1 B1 all expected numbers B1 | 2 |
| (b) | 7.42 Σ(O − E)2/E M1 7.42 A1 | 2 |
| (c) | difference not significant; accept that results fit 9 : 3 : 3 : 1 / genes assort independently 3 degrees of freedom / comparison with 7.82 B1 correct conclusion B1 | 2 |
| (d) | crossing over between non-sister chromatids of homologous chromosomes in prophase I exchanges alleles; independent assortment of homologous pairs at metaphase I; new allele combinations in gametes crossing over in prophase I between non-sister chromatids B1 chiasmata, exchange of alleles B1 random orientation of bivalents at metaphase I B1 new combinations of alleles B1 | 4 |
This question is about the control of gene expression in the bacterium E. coli.
Describe the structure of the lac operon in E. coli.
Explain how the presence of lactose leads to production of β-galactosidase.
Explain why it is an advantage to E. coli to make the enzymes coded for by the lac operon only when lactose is present.
Distinguish between structural genes and regulatory genes.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | regulatory gene for repressor; promoter; operator; structural genes for lactose permease and β-galactosidase a regulatory gene coding for a repressor protein B1 a promoter and an operator B1 structural genes lacZ (β-galactosidase) and lacY (permease) B1 | 3 |
| (b) | lactose binds repressor; repressor changes shape, leaves operator; RNA polymerase binds promoter; structural genes transcribed lactose binds to the repressor protein B1 changing its shape so it cannot bind the operator B1 RNA polymerase can bind to the promoter B1 the structural genes are transcribed and translated B1 | 4 |
| (c) | saves energy and resources avoids wasting energy and amino acids making enzymes that are not needed B2 | 2 |
| (d) | structural code for proteins with a function in the cell; regulatory code for proteins controlling other genes structural genes code for proteins used by the cell (enzymes, structural proteins) B1 regulatory genes code for proteins that control the expression of other genes B1 | 2 |
This question is about meiosis and genetic variation.
Describe the behaviour of chromosomes in prophase I and metaphase I of meiosis.
Explain how crossing over and independent assortment produce genetic variation.
Calculate the number of different combinations of chromosomes possible in human gametes from independent assortment alone.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | homologous chromosomes pair (bivalents); crossing over at chiasmata; pairs line up on equator randomly homologous chromosomes pair to form bivalents B1 non-sister chromatids cross over at chiasmata B1 bivalents line up on the equator in random orientation B1 | 3 |
| (b) | exchange of alleles between chromatids; new combinations; random orientation gives 2ⁿ combinations crossing over exchanges sections between non-sister chromatids, forming new combinations of alleles B1 independent assortment of bivalents gives 2n possible chromosome combinations in gametes B2 | 3 |
| (c) | 2²³ = 8.39 × 10⁶ 223 M1 8.39 × 106 A1 | 2 |
In a plant, purple flowers (A) are dominant to white (a), and tall stems (B) are dominant to short (b). The genes are on different chromosomes. Two plants heterozygous for both genes are crossed.
| phenotype | purple tall | purple short | white tall | white short |
|---|---|---|---|---|
| observed number | 146 | 45 | 43 | 16 |
Draw a genetic diagram to show the expected genotypes and phenotypes of the offspring, and state the expected phenotypic ratio.
Explain why the gametes have four genotypes in equal proportions.
Calculate the chi-squared value for these results.
The critical value at p = 0.05 with 3 degrees of freedom is 7.82. State what your value shows.
Explain why haemophilia is much more common in males than females.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | AaBb × AaBb; gametes AB Ab aB ab; Punnett 4×4; 9 : 3 : 3 : 1 parental genotypes AaBb × AaBb B1 gametes AB, Ab, aB, ab from each B1 16-square Punnett grid with offspring genotypes B1 ratio 9 : 3 : 3 : 1 with correct phenotypes B1 | 4 |
| (b) | independent assortment of homologous pairs in meiosis the genes are on different chromosomes B1 homologous pairs line up and separate independently in meiosis (independent assortment) B1 | 2 |
| (c) | 0.61 expected numbers 141, 46.9, 46.9, 15.6 M1 Σ(O − E)2/E M1 χ2 = 0.61 A1 | 3 |
| (d) | difference not significant; results fit 9:3:3:1 χ2 < 7.82 so the difference between observed and expected is not significant (due to chance) B1 the results support a 9 : 3 : 3 : 1 ratio B1 | 2 |
| (e) | X-linked recessive; males one X, one allele expresses; females need two the allele is recessive and on the X chromosome B1 males have only one X so a single recessive allele is expressed, females need two B1 | 2 |
In a dihybrid cross, two genes are inherited together.
Explain why the observed ratio may differ from 9 : 3 : 3 : 1.
State the purpose of a chi-squared test in genetics.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | linkage or epistasis genes may be linked on the same chromosome (or epistasis) B1 so they are not inherited independently B1 | 2 |
| (b) | whether difference between observed and expected is significant to test whether the difference between observed and expected results is significant or due to chance B2 | 2 |
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in