2022 May/June · Paper 42 · Q3 (c(i), c(ii), c(iii))8 marksCodominance, blood groups & sex linkage
2022 May/June · Paper 43 · Q2 (c, d(ii))6 marksGenes, DNA & protein synthesis · Monohybrid crosses & pedigrees
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Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Monohybrid crosses & pedigrees · Core · Paper 2 style · 1 mark
(a)Two plants that are heterozygous for flower colour (Rr) are crossed. Red (R) is dominant to white (r). What is the expected ratio of red to white offspring?[1]
Show answer
(a) B 3 : 1
Rr × Rr gives RR, Rr, Rr, rr: three red to one white. 1 : 1 (A) is the result of a test cross Rr × rr.
Q2Monohybrid crosses & pedigrees · Core · Paper 4 style · 4 marks
(a)State the meaning of (i) homozygous and (ii) heterozygous.[2]
(b)Explain why two heterozygous individuals do not 'breed true' for a characteristic.[2]
Show answer
(a) (i) Having two identical alleles of a particular gene; (ii) having two different alleles of a particular gene.
B1 each.
(b) Each can pass on either allele, so offspring can be homozygous dominant, heterozygous or homozygous recessive – some offspring show a different phenotype from the parents.
B1 gametes carry different alleles; B1 offspring can be recessive / different phenotype.
Q3Monohybrid crosses & pedigrees · Standard · Paper 4 style · 6 marks
(a)Explain how a farmer could use a test cross to find out whether a black-coated goat is homozygous (BB) or heterozygous (Bb). Black (B) is dominant to brown (b).[3]
(b)Draw a genetic diagram for the cross if the black goat is heterozygous. State the expected ratio of offspring phenotypes.[3]
Show answer
(a) Cross the black goat with a brown goat (bb, homozygous recessive); if any offspring are brown, the black goat is Bb; if all offspring (in a large number) are black, it is probably BB.
B1 cross with homozygous recessive / brown goat; B1 any brown offspring → Bb; B1 all black → BB.
(b) Bb × bb; gametes B, b and b; offspring Bb (black), bb (brown); ratio 1 black : 1 brown.
B1 parental genotypes and gametes; B1 offspring genotypes; B1 1 : 1 black : brown.
Q4Monohybrid crosses & pedigrees · Standard · Paper 6 style · 4 marks
A student in Multan grew 810 pea seeds from a cross between two tall plants. 612 plants were tall and 198 were dwarf.
(a)Calculate the ratio of tall to dwarf plants, to the nearest whole number : 1.[2]
(b)State the genotypes of the parent plants. Use T for tall and t for dwarf.[1]
(c)Suggest why the ratio is not exactly the expected ratio.[1]
Show answer
(a) 3 : 1
M1 612 ÷ 198 = 3.09; A1 3 : 1.
(b) Tt × Tt
B1 both Tt.
(c) Fertilisation is random / chance; sample is too small to give the exact ratio.
B1 random fusion of gametes / chance.
Q5Monohybrid crosses & pedigrees · Challenge · Paper 4 style · 6 marks
A family pedigree is described: Persons 1 (father) and 2 (mother) are both unaffected. They have three children: person 3 (son, unaffected), person 4 (daughter, unaffected) and person 5 (son, has cystic fibrosis).
(a)Cystic fibrosis is caused by a recessive allele (f). Use the pedigree to state the genotypes of persons 1 and 2. Explain your answer.[3]
(b)Calculate the probability that the next child of persons 1 and 2 will have cystic fibrosis.[1]
(c)Person 4 is unaffected. Calculate the probability that person 4 is a carrier (Ff).[2]
Show answer
(a) Both Ff. They are unaffected, so each has at least one F; their child 5 has cystic fibrosis (ff) and received one f from each parent.
B1 both Ff; B1 unaffected so have dominant allele; B1 affected child must have received f from each.
(b) 0.25
B1 1/4 = 0.25.
(c) 2/3 (0.67)
M1 unaffected offspring of Ff × Ff are FF, Ff, Ff (ff excluded); A1 2/3.
Q6Genes, DNA & protein synthesis · Core · Paper 2 style · 1 mark
(a)Where in a cell are proteins assembled from amino acids?[1]
Show answer
(a) B ribosome
mRNA carries a copy of the gene from the nucleus to a ribosome in the cytoplasm, where amino acids are joined in order. The nucleus (A) is where the mRNA copy is made, not the protein.
Q7Genes, DNA & protein synthesis · Core · Paper 4 style · 5 marks
(a)Define the terms (i) gene and (ii) allele.[2]
(b)State what a chromosome is made of.[1]
(c)State the number of chromosomes in (i) a human skin cell and (ii) a human egg cell.[2]
Show answer
(a) (i) A length of DNA that codes for a protein; (ii) an alternative form of a gene.
B1 each.
(b) DNA (carrying genetic information in the form of genes).
B1 DNA.
(c) (i) 46; (ii) 23.
B1 each.
Q8Genes, DNA & protein synthesis · Core · Paper 4 style · 3 marks
(a)Name the molecule that carries a copy of a gene from the nucleus to the cytoplasm.[1]
(b)Name the units that are joined together to make a protein.[1]
(c)State what determines the order in which these units are joined.[1]
Show answer
(a) mRNA (messenger RNA)
B1 mRNA.
(b) Amino acids
B1 amino acids.
(c) The sequence of bases in the gene / mRNA.
B1 base sequence.
Q9Genes, DNA & protein synthesis · Standard · Paper 4 style · 5 marks
The gene for insulin is found in all human body cells but is used only in some cells of the pancreas.
(a)Describe how the information in a gene is used to make a protein.[5]
Show answer
(a) The gene is a sequence of bases in DNA in the nucleus; a copy of the gene is made as an mRNA molecule; mRNA moves out of the nucleus into the cytoplasm; mRNA attaches to a ribosome; the ribosome assembles amino acids into a protein molecule; the specific order of bases in the mRNA determines the specific sequence of amino acids.
1 mark each for any five of: DNA stays in nucleus; mRNA made as a copy of gene; mRNA leaves nucleus (through pores) to cytoplasm; mRNA attaches to ribosome; ribosome joins amino acids; base sequence determines amino acid sequence; amino acid sequence determines the protein's shape / function.
Q10Genes, DNA & protein synthesis · Standard · Paper 4 style · 5 marks
(a)A gene contains 1 260 bases on its coding strand. Three bases code for one amino acid. Assuming every base in the gene codes for an amino acid, calculate the number of amino acids in the protein it codes for.[1]
(b)A mutation removes one base near the start of the gene. Suggest why this usually has a large effect on the protein made.[2]
(c)State why the shape of a protein is important for its function, using an enzyme as an example.[2]
Show answer
(a) 420
B1 1260 ÷ 3 = 420.
(b) The bases after the deletion are read in different groups of three, so the sequence of amino acids changes from that point; the protein has a different shape and may not function.
B1 changes the reading of all later bases / amino acid sequence; B1 protein shape / function changed.
(c) The amino acid sequence determines how the protein folds; an enzyme's active site must have a specific shape complementary to its substrate.
B1 sequence determines shape; B1 active site shape complementary to substrate.
Q11Genes, DNA & protein synthesis · Standard · Paper 4 style · 4 marks
(a)Explain what is meant by the statement 'a person's genotype for eye colour is homozygous recessive'.[2]
(b)Define (i) phenotype and (ii) genotype.[2]
Show answer
(a) Homozygous: has two identical alleles of the gene; recessive: the allele only expressed when no dominant allele is present (e.g. bb).
B1 two identical alleles; B1 recessive alleles (only expressed when homozygous).
(b) (i) The observable features of an organism; (ii) the genetic make-up of an organism in terms of the alleles present.
B1 each.
Q12Genes, DNA & protein synthesis · Challenge · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) B Different genes are expressed (switched on) in each type of cell.
All body cells carry the same genes, but each specialised cell expresses only the genes for the proteins it needs. Body cells do not lose genes (C) and have the same chromosome number (A).
Q13Genes, DNA & protein synthesis · Challenge · Paper 4 style · 4 marks
(a)Explain why a change in the base sequence of a gene can result in a protein that does not work.[3]
(b)Most body cells do not express the gene for haemoglobin, though they contain it. Suggest one advantage of switching off unneeded genes.[1]
Show answer
(a) The base sequence determines the sequence of amino acids; a different amino acid sequence changes how the protein folds (its shape); e.g. an enzyme's active site no longer fits its substrate.
B1 base sequence → amino acid sequence; B1 changed shape; B1 consequence for function, e.g. active site / binding site.
(b) Saves energy and materials (amino acids) by not making proteins the cell does not need.
B1 saves energy / resources; accept prevents proteins interfering with the cell's own function.
Q14Genes, DNA & protein synthesis · Challenge · Paper 4 style · 4 marks
(a)A protein is made of 285 amino acids. Calculate the minimum number of bases in the mRNA that codes for it, given three bases code for one amino acid.[1]
(b)The gene in the DNA is longer than this. Suggest one reason.[1]
(c)Explain why DNA does not leave the nucleus to make proteins, but mRNA does.[2]
Show answer
(a) 855
B1 285 × 3 = 855.
(b) Not all of the DNA of a gene codes for amino acids, e.g. start/stop signals or regions that are removed / do not code.
B1 non-coding regions / stop signal.
(c) DNA is very large / is the master copy, which must be protected from damage in the cytoplasm; mRNA is a small, short-lived copy of one gene that can pass through the nuclear pores.
B1 DNA too large / kept safe; B1 mRNA small copy of a single gene that can leave nucleus.
Q15Codominance, blood groups & sex linkage · Core · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) A A or B only
Father IᴬIᴮ gives Iᴬ or Iᴮ; mother IᴼIᴼ gives Iᴼ; children are IᴬIᴼ (group A) or IᴮIᴼ (group B). No child can be AB or O.
Q16Codominance, blood groups & sex linkage · Core · Paper 6 style · 4 marks
A class of 40 students in Hyderabad recorded their blood groups: A – 9, B – 13, AB – 4, O – 14.
(a)Calculate the percentage of the 40 students with blood group O.[1]
(b)State which blood group shows that two alleles are codominant.[1]
(c)Suggest a suitable way of displaying these results and give a reason.[2]
Show answer
(a) 35 %
B1 14 ÷ 40 × 100 = 35 %.
(b) AB
B1 AB.
(c) A bar chart (with gaps between bars), because blood group is a discontinuous (categoric) variable.
B1 bar chart; B1 discontinuous / categories.
Q17Codominance, blood groups & sex linkage · Standard · Paper 2 style · 1 mark
(a)Red-green colour blindness is caused by a recessive allele on the X chromosome. A man with normal vision and a woman who is a carrier have a son. What is the probability that the son is colour blind?[1]
Show answer
(a) C 0.5
A son gets Y from his father and one X from his mother: Xᴮ or Xᵇ with equal chance, so 0.5. 0.25 (B) is the probability for any child (son or daughter) being colour blind.
Q18Codominance, blood groups & sex linkage · Standard · Paper 4 style · 7 marks
(a)Explain what is meant by codominance, using the ABO blood group alleles as an example.[2]
(b)Draw a genetic diagram to show the possible blood groups of children from a father who is IᴬIᴼ and a mother who is IᴮIᴼ. Give the probability of a child having blood group O.[5]
Show answer
(a) Both alleles in a heterozygote are expressed in the phenotype; e.g. Iᴬ and Iᴮ are codominant so a person with IᴬIᴮ has blood group AB.
B1 both alleles expressed in heterozygote; B1 IᴬIᴮ gives AB.
(b) Parental genotypes IᴬIᴼ × IᴮIᴼ; gametes Iᴬ, Iᴼ and Iᴮ, Iᴼ; offspring IᴬIᴮ (AB), IᴬIᴼ (A), IᴮIᴼ (B), IᴼIᴼ (O); probability of O = 0.25 (1 in 4).
Q19Codominance, blood groups & sex linkage · Standard · Paper 4 style · 6 marks
(a)State the blood group of a person with each genotype: (i) IᴮIᴼ, (ii) IᴬIᴮ, (iii) IᴼIᴼ.[3]
(b)A couple both have blood group A. Their first child has blood group O. State the genotypes of the parents and the probability that their next child will have blood group A.[3]
Show answer
(a) (i) B; (ii) AB; (iii) O.
B1 each.
(b) Both parents IᴬIᴼ; probability of group A = 0.75.
(c) All eggs carry an X chromosome; half of sperm carry X and half carry Y.
B1 eggs all X, sperm X or Y.
Q21Codominance, blood groups & sex linkage · Challenge · Paper 4 style · 7 marks
(a)Explain why red-green colour blindness is more common in men than in women.[3]
(b)A colour-blind man (XᵇY) and a woman with normal vision who is not a carrier (XᴮXᴮ) have children. Use a genetic diagram to give the probability that (i) a daughter is a carrier and (ii) a son is colour blind.[4]
Show answer
(a) The allele is on the X chromosome (sex-linked) and the Y chromosome carries no copy of the gene; a man has only one X, so one recessive allele (XᵇY) makes him colour blind; a woman needs two recessive alleles (XᵇXᵇ), which is less likely.
B1 gene on X chromosome / not on Y; B1 males need only one recessive allele; B1 females need two / heterozygous females are carriers.
(b) Gametes Xᵇ, Y and Xᴮ; daughters XᴮXᵇ, sons XᴮY. (i) 1 (all daughters carriers); (ii) 0.
B1 parental genotypes and gametes; B1 offspring genotypes; B1 (i) 1 / 100 %; B1 (ii) 0.
Q22Codominance, blood groups & sex linkage · Challenge · Paper 4 style · 5 marks
(a)A woman whose father was colour blind has normal vision. She has children with a man who has normal vision. Give the woman's genotype and explain how you worked it out.[2]
(b)Calculate the probability that their first child is a colour-blind boy.[2]
(c)Calculate the probability that two of their children are both colour-blind boys.[1]
Show answer
(a) XᴮXᵇ; she has normal vision so has at least one Xᴮ, and she must have received Xᵇ from her colour-blind father (fathers pass their X to daughters).
B1 XᴮXᵇ; B1 reasoning – father gives his only X (Xᵇ) to daughter.
(b) 0.25
M1 offspring XᴮXᴮ, XᴮXᵇ, XᴮY, XᵇY; A1 1/4 = 0.25.
(c) 0.0625 (1/16)
B1 0.25 × 0.25 = 0.0625.
Q23Mitosis, meiosis & stem cells · Core · Paper 2 style · 1 mark
(a)How many chromosomes are present in each cell produced when a human body cell (46 chromosomes) divides by mitosis?[1]
Show answer
(a) B 46
Mitosis produces two genetically identical diploid cells with the same chromosome number as the parent, 46. 23 (A) is the result of meiosis.
Q24Mitosis, meiosis & stem cells · Core · Paper 4 style · 4 marks
(a)State three roles of mitosis in the human body.[3]
(b)State the type of cell division that produces sperm cells.[1]
Show answer
(a) Growth; repair of damaged tissues; replacement of cells (e.g. skin, red blood cells).
B1 each; accept asexual reproduction only for other organisms.
(b) Meiosis
B1 meiosis.
Q25Mitosis, meiosis & stem cells · Core · Paper 4 style · 4 marks
(a)Define (i) a haploid nucleus and (ii) a diploid nucleus.[2]
(b)Give one example of a haploid cell and one example of a diploid cell in a human.[2]
Show answer
(a) (i) A nucleus containing a single set of chromosomes; (ii) a nucleus containing two sets of chromosomes.
B1 each.
(b) Haploid: sperm or egg cell; diploid: any body cell, e.g. a cheek cell or zygote.
B1 each.
Q26Mitosis, meiosis & stem cells · Standard · Paper 4 style · 4 marks
A patient in a hospital in Lahore needs treatment to replace damaged blood-forming tissue.
(a)Define the term stem cell.[2]
(b)Explain why stem cells taken from a patient's own bone marrow can be used to replace damaged cells without being rejected.[2]
Show answer
(a) An unspecialised cell that divides by mitosis to produce daughter cells that can become specialised for specific functions.
B1 unspecialised; B1 divides by mitosis to give cells that can become specialised.
(b) The new cells are genetically identical to the patient's (made by mitosis), so they have the same antigens and are not recognised as foreign by the immune system.
B1 genetically identical / same DNA; B1 same antigens / not attacked by immune system.
Q27Mitosis, meiosis & stem cells · Standard · Paper 6 style · 4 marks
A student examined a stained squash of an onion root tip under a microscope. In one field of view 150 cells were counted and 24 of them showed chromosomes that were visible (dividing).
(a)Calculate the percentage of cells that were dividing (mitotic index).[2]
(b)Suggest why the student used the tip of the root rather than a region further back.[1]
(c)Suggest how the student could make the estimate more reliable.[1]
Show answer
(a) 16 %
M1 (24 ÷ 150) × 100; A1 16 %.
(b) Cells divide by mitosis mainly at the root tip (growth region); further back cells are elongating / specialising.
B1 mitosis occurs at tip / growth region.
(c) Count cells in more fields of view / from more roots and calculate a mean.
B1 more fields / more roots / repeats.
Q28Mitosis, meiosis & stem cells · Standard · Paper 4 style · 5 marks
(a)Compare mitosis and meiosis by giving three differences.[3]
(b)Mitosis is also used in asexual reproduction. Explain why a field of potato plants grown from tubers of one plant all show the same disease susceptibility.[2]
Show answer
(a) Mitosis gives 2 cells, meiosis gives 4 (cells / gametes); mitosis gives genetically identical cells, meiosis gives genetically different cells; mitosis keeps chromosome number (diploid), meiosis halves it (haploid); mitosis for growth/repair, meiosis for gamete production.
1 mark each for any three comparative differences.
(b) Tubers are produced by mitosis so all the plants are genetically identical (clones); none has alleles giving resistance, so all are equally affected.
B1 mitosis → genetically identical / clones; B1 same alleles / no variation for resistance.
Q29Mitosis, meiosis & stem cells · Challenge · Paper 4 style · 6 marks
(a)Describe what happens to chromosomes before and during mitosis so that the two daughter cells are genetically identical.[3]
(b)Explain why it is important that gametes are produced by meiosis, not by mitosis.[3]
Show answer
(a) Each chromosome is copied (replicated / DNA copied) before division; the two copies separate, one copy going to each end of the cell; each new nucleus receives an exact copy of every chromosome; cell then divides (cytoplasm).
B1 chromosomes replicate before mitosis; B1 copies separate (to opposite poles); B1 each daughter nucleus gets full identical set.
(b) Meiosis halves the chromosome number (diploid 46 → haploid 23); at fertilisation two gametes fuse to restore the diploid number; otherwise the chromosome number would double every generation; meiosis also produces genetic variation.
B1 halves chromosome number / haploid; B1 fertilisation restores diploid; B1 prevents doubling each generation / gives variation.
Q30Mitosis, meiosis & stem cells · Challenge · Paper 4 style · 5 marks
(a)Calculate how many cells would be produced from one fertilised egg after 12 rounds of mitosis, assuming every cell divides each time.[1]
(b)Red blood cells have no nucleus and live for only about four months. Explain why stem cells in the bone marrow must keep dividing throughout a person's life.[2]
(c)State the chromosome number in the egg of a dog, whose body cells have 78 chromosomes, and name the division producing it.[2]
Show answer
(a) 4096
B1 2¹² = 4096.
(b) Red blood cells cannot divide (no nucleus) and are constantly being destroyed, so they must be replaced; stem cells divide by mitosis to make new cells that become specialised as red blood cells.
B1 red blood cells cannot divide / are continually lost and need replacing; B1 stem cells divide by mitosis and the daughter cells become specialised (as red blood cells).
(c) 39; meiosis.
B1 39; B1 meiosis.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
