Secondary 4 Pure Biology Genetics Inheritance Quiz

Secondary 4 Pure Biology Quiz - Genetics Inheritance

ANSWER KEY AND MARKING SCHEME

Total Marks: 40

Section A: Multiple Choice (5 marks)

Marking Notes:

• Q1: C is correct (A-T, C-G base pairing in DNA). A is incorrect (wrong pairing). B is incorrect (wrong pairing). D is incorrect (Uracil is in RNA, not DNA).
• Q2: A is correct. Man: I^A I^B, Woman: I^O I^O. Children can be I^A I^O (blood group A) or I^B I^O (blood group B) only.
• Q3: B is correct. Tt × Tt → 1 TT : 2 Tt : 1 tt, giving 3 tall : 1 dwarf phenotype.
• Q4: B is correct. A gene is a segment of DNA that codes for one polypeptide.
• Q5: C is correct. Mother: X^C X^c, Father: X^c Y. Sons inherit Y from father and either X^C or X^c from mother. 50% chance of X^c Y (colour-blind son).

Section B: Short Answer (15 marks)

6. Distinguish between genotype and phenotype. [2 marks]

Answer:

• Genotype refers to the genetic makeup / combination of alleles an organism possesses for a particular trait.
• Phenotype refers to the observable / expressed characteristics or traits of an organism, resulting from the interaction of its genotype with the environment.

Marking:

• 1 mark for correct definition of genotype (must mention genetic makeup or alleles).
• 1 mark for correct definition of phenotype (must mention observable characteristics or expression).
• Accept: "Genotype is the set of genes; phenotype is the physical appearance."

7. State what is meant by the term homozygous. [1 mark]

Answer: An organism is homozygous for a trait when it possesses two identical alleles for that gene on homologous chromosomes (e.g., TT or tt).

Marking:

• 1 mark for "two identical alleles" or equivalent phrasing.
• Accept: "Having the same alleles for a particular gene."

8. Explain why a recessive allele is only expressed in the phenotype when present in the homozygous condition. [2 marks]

Answer: A recessive allele codes for a non-functional or less functional protein/enzyme. In a heterozygous individual, the dominant allele produces sufficient functional protein to mask the effect of the recessive allele. The recessive phenotype only appears when both alleles are recessive (homozygous recessive), so no functional protein is produced.

Marking:

• 1 mark for explaining that the dominant allele masks the recessive allele in heterozygous condition (or that dominant allele produces functional protein).
• 1 mark for stating that recessive phenotype requires two recessive alleles (homozygous recessive) / no dominant allele present.
• Accept reference to enzyme/protein function or lack thereof.

9. Diagram of homologous chromosomes. [2 marks]

(a) State the genotype of the individual for gene A. [1 mark]

Answer: Aa (heterozygous)

Marking:

• 1 mark for "Aa" or "heterozygous."
• Do not accept just "A" or "a."

(b) Explain what is meant by the term allele. [1 mark]

Answer: An allele is an alternative form of a gene found at the same locus/position on homologous chromosomes.

Marking:

• 1 mark for "alternative form of a gene" or "different version of a gene."
• Accept reference to same locus/position.

10. Guinea pig coat colour inheritance. [4 marks]

(a) State the genotype of the F₁ generation. [1 mark]

Answer: Bb (all heterozygous black)

Marking:

• 1 mark for "Bb" or "heterozygous."

(b) Genetic diagram for F₂ generation. [3 marks]

Answer: Parental phenotypes: Black coat × Black coat
Parental genotypes: Bb × Bb
Gametes: B, b and B, b

F₂ genotypic ratio: 1 BB : 2 Bb : 1 bb
F₂ phenotypic ratio: 3 black coat : 1 white coat

Marking:

• 1 mark for correct parental genotypes (Bb × Bb) and gametes (B, b from each).
• 1 mark for correct Punnett square or equivalent working.
• 1 mark for correct phenotypic ratio (3:1) with phenotypes stated.
• Accept equivalent genetic diagram format (e.g., branching method).

Section C: Structured Questions (10 marks)

11. Explain codominance using ABO blood groups. [2 marks]

Answer: Codominance occurs when both alleles in a heterozygous individual are fully expressed in the phenotype, with neither allele being dominant over the other. In ABO blood groups, the alleles I^A and I^B are codominant. A person with genotype I^A I^B has blood group AB, expressing both A and B antigens on their red blood cells.

Marking:

• 1 mark for definition: both alleles expressed equally in heterozygous condition.
• 1 mark for correct ABO example: I^A I^B produces blood group AB with both antigens.
• Accept reference to both antigens being present.

12. State two differences between mitosis and meiosis. [2 marks]

Answer:

1. Mitosis produces two genetically identical daughter cells; meiosis produces four genetically different daughter cells.
2. Mitosis involves one division; meiosis involves two divisions.
3. Mitosis produces diploid cells (same chromosome number as parent); meiosis produces haploid cells (half the chromosome number).
4. Mitosis occurs in body cells for growth and repair; meiosis occurs in reproductive organs to produce gametes.

Marking:

• 1 mark for each correct difference (any two from above or equivalent).
• Must be clear comparison (e.g., "mitosis produces 2 cells, meiosis produces 4 cells").
• Do not award marks for vague statements without comparison.

13. Pedigree analysis. [8 marks]

(a) State whether the disorder is dominant or recessive. Explain. [2 marks]

Answer: The disorder is caused by a recessive allele.
Explanation: Unaffected parents (I-1 and I-2) have an affected child (II-3). If the disorder were dominant, at least one parent would need to be affected. The fact that two unaffected parents can produce an affected child indicates the disorder is recessive, and both parents are carriers (heterozygous).

Marking:

• 1 mark for "recessive."
• 1 mark for correct explanation referencing unaffected parents producing affected child / skipping generations.
• Accept: "Parents I-1 and I-2 are unaffected but have affected son II-3, showing the trait is recessive."

(b) Genotype of individual II-3. Explain. [2 marks]

Answer: Genotype: dd (homozygous recessive)
Explanation: Individual II-3 is affected by the recessive disorder, so he must have inherited one recessive allele from each parent. Since the disorder is recessive, affected individuals must be homozygous recessive.

Marking:

• 1 mark for "dd" or "homozygous recessive."
• 1 mark for explanation: affected by recessive disorder → must have two recessive alleles.

(c) Genetic diagram for probability of affected child. [4 marks]

Answer: Individual III-2 is affected → genotype dd.
His wife is unaffected but her father was affected (dd) → she must be a carrier → genotype Dd.

Parental phenotypes: Affected male × Unaffected carrier female
Parental genotypes: dd × Dd
Gametes: d, d and D, d

Offspring genotypes: 2 Dd : 2 dd
Offspring phenotypes: 2 unaffected carriers : 2 affected
Probability of affected child = 2/4 = 50% or 1/2

Marking:

• 1 mark for correctly deducing wife's genotype as Dd (carrier) with reasoning.
• 1 mark for correct parental genotypes (dd × Dd) and gametes.
• 1 mark for correct Punnett square or equivalent.
• 1 mark for correct probability (50% or 1/2).
• Accept equivalent working; deduct 1 mark if probability not stated.

14. DNA structure. [5 marks]

(a) Name the sugar molecule labelled S. [1 mark]

Answer: Deoxyribose (sugar)

Marking:

• 1 mark for "deoxyribose."
• Do not accept "ribose" (that is RNA sugar).

(b) Type of bond between complementary bases. [1 mark]

Answer: Hydrogen bond(s)

Marking:

• 1 mark for "hydrogen bond(s)."
• Accept "weak hydrogen bonds."

(c) Explain how DNA structure allows accurate replication. [2 marks]

Answer: DNA is a double helix with two complementary strands. Each strand acts as a template for the synthesis of a new complementary strand. The specific base pairing rules (A with T, C with G) ensure that the new strand is an exact copy of the original complementary strand, resulting in two identical DNA molecules.

Marking:

• 1 mark for mentioning complementary base pairing / specific base pairing rules (A-T, C-G).
• 1 mark for explaining that each strand serves as a template for replication.
• Accept reference to semi-conservative replication.

(d) Complementary base sequence. [1 mark]

Answer: T A C G G A T C

Marking:

• 1 mark for correct sequence: T A C G G A T C.
• Must be written in correct order.

15. Genetic engineering – human insulin. [7 marks]

(a) Steps in producing human insulin. [4 marks]

Answer:

1. Identify and isolate the human gene for insulin production.
2. Use restriction enzymes to cut the human DNA and a plasmid (vector) from bacteria, creating sticky ends.
3. Insert the human insulin gene into the plasmid using DNA ligase to form recombinant DNA.
4. Introduce the recombinant plasmid into host bacteria (transformation).
5. Culture the transgenic bacteria in fermenters, where they multiply and express the human insulin gene to produce insulin.
6. Extract and purify the human insulin for medical use.

Marking:

• 1 mark for isolation of insulin gene.
• 1 mark for use of restriction enzymes and plasmid/vector.
• 1 mark for insertion using DNA ligase / formation of recombinant DNA.
• 1 mark for introduction into bacteria and production/extraction.
• Accept any four clear, sequential steps.

(b) One benefit and one ethical concern. [3 marks]

Answer: Benefit: Human insulin produced by bacteria is identical to natural human insulin, reducing the risk of allergic reactions compared to animal insulin (e.g., from pigs/cows). It can also be produced in large quantities to meet demand. Ethical concern: Some people object to the manipulation of genetic material, considering it "playing God" or interfering with nature. There are also concerns about the long-term effects and safety of genetically modified organisms.

Marking:

• 1 mark for a valid benefit with explanation.
• 1 mark for a valid ethical concern with explanation.
• 1 mark for clarity and relevance of discussion.
• Accept other valid benefits (e.g., cheaper production, consistent quality) and ethical concerns (e.g., animal welfare, patenting of genes).

Section D: Extended Questions (10 marks)

16. Protein synthesis. [5 marks]

Answer: Transcription:

• The DNA double helix unwinds and unzips at the gene region.
• One strand acts as a template.
• RNA polymerase synthesises a complementary mRNA strand using free RNA nucleotides, following base pairing rules (A-U, C-G).
• The mRNA strand detaches and moves out of the nucleus to the ribosome.

Translation:

• mRNA attaches to a ribosome.
• tRNA molecules with specific anticodons bring amino acids to the ribosome.
• The tRNA anticodon pairs with the complementary mRNA codon.
• Amino acids are joined by peptide bonds to form a polypeptide chain.
• The process continues until a stop codon is reached, and the polypeptide is released.

Marking:

• 1 mark for unwinding of DNA and template strand.
• 1 mark for mRNA synthesis by RNA polymerase / base pairing.
• 1 mark for mRNA moving to ribosome.
• 1 mark for role of tRNA, anticodons, and codons.
• 1 mark for formation of polypeptide chain / peptide bonds.
• Accept clear, labelled diagrams.

17. Mutations and genetic variation. [5 marks]

Answer: Mutations are changes in the DNA sequence or chromosome structure/number. They are a source of new alleles and genetic variation.

Gene mutations:

• A change in the nucleotide sequence of a single gene.
• Example: Sickle cell anaemia, caused by a substitution mutation in the haemoglobin gene, leading to a different amino acid and altered protein shape.

Chromosomal mutations:

• Changes in the structure or number of entire chromosomes.
• Example: Down syndrome, caused by trisomy 21 (an extra copy of chromosome 21).

Mutations create new alleles that may produce different phenotypes, contributing to variation within a population upon which natural selection can act.

Marking:

• 1 mark for definition of mutation.
• 1 mark for distinction between gene and chromosomal mutations.
• 1 mark for a correct example of a gene mutation.
• 1 mark for a correct example of a chromosomal mutation.
• 1 mark for linking mutations to new alleles/genetic variation.

18. Incomplete dominance. [5 marks]

(a) Explanation of pink flowers. [2 marks]

Answer: Incomplete dominance occurs when neither allele is completely dominant over the other. The heterozygous genotype (Rr) produces an intermediate phenotype (pink flowers) because the single functional allele produces only half the amount of pigment compared to the homozygous dominant (RR), resulting in a blend of red and white.

Marking:

• 1 mark for defining incomplete dominance / neither allele dominant.
• 1 mark for explaining intermediate phenotype in heterozygote.

(b) Genetic diagram for F₂ generation. [3 marks]

Answer: Parental phenotypes: Pink flowers × Pink flowers
Parental genotypes: Rr × Rr
Gametes: R, r and R, r

F₂ genotypic ratio: 1 RR : 2 Rr : 1 rr
F₂ phenotypic ratio: 1 red : 2 pink : 1 white

Marking:

• 1 mark for correct parental genotypes and gametes.
• 1 mark for correct Punnett square.
• 1 mark for correct phenotypic ratio (1:2:1) with phenotypes stated.

19. Meiosis and genetic variation. [5 marks]

Answer: Meiosis produces genetic variation through two main mechanisms:

1. Crossing over: During prophase I, homologous chromosomes pair up and exchange segments of DNA. This creates new combinations of alleles on a chromosome (recombinant chromosomes).
2. Independent assortment: During metaphase I, homologous pairs line up randomly at the equator. The orientation of each pair is independent of others, leading to different combinations of maternal and paternal chromosomes in the gametes.
3. Random fertilisation: The fusion of unique male and female gametes further increases genetic variation in the offspring.

These processes ensure that offspring are genetically different from parents and siblings, which is important for adaptation and evolution.

Marking:

• 1 mark for stating meiosis produces genetic variation.
• 1 mark for explaining crossing over with detail.
• 1 mark for explaining independent assortment with detail.
• 1 mark for mentioning random fertilisation.
• 1 mark for linking variation to evolution/adaptation.

20. GM crops benefits and risks. [5 marks]

Answer: Benefits:

• Increased crop yield and food security by introducing traits like pest resistance (e.g., Bt corn) or herbicide tolerance.
• Enhanced nutritional content (e.g., Golden Rice with beta-carotene to combat vitamin A deficiency).
• Reduced use of chemical pesticides, lowering environmental impact and costs.

Risks:

• Potential for GM genes to spread to wild relatives (gene flow), creating superweeds.
• Possible allergic reactions or unknown long-term health effects in humans.
• Ethical concerns about corporate control of seed supply and impact on small farmers.
• Reduction in biodiversity due to monoculture practices.

Marking:

• 1 mark for a valid benefit with explanation.
• 1 mark for a second benefit or elaboration.
• 1 mark for a valid risk with explanation.
• 1 mark for a second risk or elaboration.
• 1 mark for balanced evaluation/conclusion.
• Accept other valid points; must show evaluation, not just listing.

END OF ANSWER KEY