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Secondary 3 Biology Genetics Inheritance Quiz
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Questions
Secondary 3 Biology Quiz - Genetics Inheritance
Name: _________________________ Class: _________________________ Date: _________________________ Score: ________ / 40
Duration: 45 minutes Total Marks: 40
Instructions:
- Answer ALL questions in the spaces provided.
- Write your answers clearly and legibly.
- Marks are indicated in brackets [ ] at the end of each question or part-question.
- For multiple choice questions, circle the letter of the correct answer.
- For structured questions, use precise biological terminology.
Section A: Multiple Choice (5 marks)
Circle the correct answer for each question.
1. Which of the following correctly describes the relationship between DNA, genes, and chromosomes? A) Genes are made of chromosomes, which are made of DNA B) DNA is made of genes, which are made of chromosomes C) Chromosomes are made of DNA, which contains genes D) Chromosomes are made of genes, which contain DNA [1]
2. In a DNA molecule, which base pairs with adenine (A)? A) Adenine B) Guanine C) Cytosine D) Thymine [1]
3. A gene is a section of DNA that codes for: A) One amino acid B) One polypeptide C) One chromosome D) One nucleotide [1]
4. Which of the following represents a homozygous recessive genotype? A) AA B) Aa C) aa D) AB [1]
5. In humans, the sex chromosomes of a female are: A) XY B) XX C) YY D) XO [1]
Section B: Short Answer (10 marks)
Answer each question in the space provided.
6. Define the term "allele."
[2]
7. Distinguish between the terms "genotype" and "phenotype."
[2]
8. State the number of chromosomes found in a normal human: (a) Body cell (somatic cell): __________ (b) Gamete (sex cell): __________ [2]
9. Explain what is meant by "complementary base pairing" in DNA.
[2]
10. A heterozygous tall pea plant (Tt) is crossed with a homozygous dwarf pea plant (tt). Using the symbols T = tall (dominant) and t = dwarf (recessive), state the expected phenotypic ratio of the offspring.
[2]
Section C: Structured Questions (15 marks)
Answer all parts of each question.
11. The diagram below shows a short section of a DNA molecule.
[Diagram showing a simplified DNA double helix with labels A, B, C, and D pointing to different components]
(a) Identify the structures labelled A, B, C, and D. A: ________________________ B: ________________________ C: ________________________ D: ________________________ [4]
(b) Explain how the structure of DNA enables it to replicate accurately.
[3]
12. Sickle cell anaemia is an inherited disease caused by a recessive allele. Individuals who are homozygous recessive suffer from the disease. The pedigree diagram below shows how sickle cell anaemia was inherited in one family.
[Pedigree diagram showing:
Generation I: Unaffected male (square, empty) married to unaffected female (circle, empty)
Generation II: Three children - unaffected male (square, empty), affected female (circle, filled), unaffected female (circle, empty)
Generation II affected female married to unaffected male (square, empty)
Generation III: Two children - affected male (square, filled), unaffected female (circle, empty)]
(a) Using the symbol H for the normal allele and h for the sickle cell allele, state the genotypes of: (i) The affected female in Generation II: __________ (ii) Her husband (the unaffected male she married): __________ [2]
(b) Explain why the affected female's parents (Generation I) are both unaffected but were able to have an affected child.
[3]
(c) The affected female in Generation II and her husband are expecting another child. Using a genetic diagram (Punnett square), determine the probability that this child will have sickle cell anaemia.
Probability: __________ [3]
13. A scientist investigated the inheritance of flower colour in a species of plant. She crossed a pure-breeding red-flowered plant with a pure-breeding white-flowered plant. All the offspring (F₁ generation) had red flowers. She then allowed the F₁ plants to self-pollinate and obtained 600 F₂ offspring.
The results of the F₂ generation are shown in the table below:
| Flower Colour | Number of Plants |
|---|---|
| Red | 452 |
| White | 148 |
(a) Using the symbols R for the dominant allele and r for the recessive allele, state the genotypes of: (i) The pure-breeding red-flowered parent: __________ (ii) The pure-breeding white-flowered parent: __________ (iii) The F₁ offspring: __________ [3]
(b) Explain why all the F₁ offspring had red flowers.
[2]
(c) Calculate the expected ratio of red-flowered to white-flowered plants in the F₂ generation based on Mendelian inheritance.
Expected ratio: __________ [1]
(d) The observed results show 452 red and 148 white plants. Explain whether these results are consistent with the expected Mendelian ratio.
[4]
14. Describe the difference between a dominant allele and a recessive allele.
[2]
15. A man with blood type A (genotype IAi) marries a woman with blood type B (genotype IBi). Using a Punnett square, determine the possible blood types of their children and the probability of each type.
Probability of each blood type:
[3]
Section D: Data-Based and Application Questions (10 marks)
Answer all parts of each question.
16. In pea plants, round seeds (R) are dominant over wrinkled seeds (r). A heterozygous round-seeded plant is crossed with a wrinkled-seeded plant.
(a) State the genotypes of the parent plants.
[1]
(b) Draw a Punnett square to show the possible genotypes of the offspring. [2]
(c) State the phenotypic ratio of the offspring.
[1]
17. Explain how a mutation in a gene can lead to a change in the protein produced.
[3]
18. A couple has three sons. The woman is pregnant with a fourth child. What is the probability that this child will be a boy? Explain your answer.
[2]
19. Define the term "homozygous" and give an example using the symbols A and a.
[1]
20. Explain one advantage and one disadvantage of genetic screening for inherited diseases. Advantage: ________________________________________________________________________ Disadvantage: ________________________________________________________________________ [2]
END OF QUIZ
Answers
Secondary 3 Biology Quiz - Genetics Inheritance - ANSWER KEY
Total Marks: 40
Section A: Multiple Choice (5 marks)
1. C) Chromosomes are made of DNA, which contains genes [1]
- Marking note: Accept C only. Chromosomes are structures composed of DNA; genes are specific sequences of DNA on chromosomes.
2. D) Thymine [1]
- Marking note: Adenine pairs with thymine (A-T); cytosine pairs with guanine (C-G).
3. B) One polypeptide [1]
- Marking note: Each gene codes for one polypeptide (or one protein). A gene contains many nucleotides and codes for many amino acids, but the product is one polypeptide chain.
4. C) aa [1]
- Marking note: Homozygous means both alleles are the same; recessive means the allele is expressed only when homozygous. "aa" represents homozygous recessive.
5. B) XX [1]
- Marking note: Females have two X chromosomes (XX); males have one X and one Y (XY).
Section B: Short Answer (10 marks)
6. An allele is an alternative form of a gene that occupies the same position (locus) on homologous chromosomes and controls the same characteristic. [2]
- Marking note: Award 1 mark for "alternative form of a gene" and 1 mark for "same position/locus on homologous chromosomes" or equivalent.
7. Genotype refers to the genetic makeup of an organism (the combination of alleles it possesses for a particular trait). Phenotype refers to the observable physical or physiological characteristics of an organism, which result from the interaction of its genotype with the environment. [2]
- Marking note: Award 1 mark for correct definition of genotype (genetic makeup/allele combination) and 1 mark for correct definition of phenotype (observable characteristics/expression of genotype).
8. (a) 46 [1] (b) 23 [1]
- Marking note: Body cells are diploid (2n = 46); gametes are haploid (n = 23). Accept "46 chromosomes / 23 pairs" for (a) and "23 chromosomes / half the number" for (b).
9. Complementary base pairing refers to the specific pairing of nitrogenous bases in DNA where adenine (A) always pairs with thymine (T) via two hydrogen bonds, and cytosine (C) always pairs with guanine (G) via three hydrogen bonds. This ensures accurate replication and maintains the uniform width of the DNA double helix. [2]
- Marking note: Award 1 mark for stating A-T and C-G pairing, and 1 mark for explaining that this ensures accurate replication or maintains DNA structure. Accept mention of hydrogen bonds.
10. 1 tall : 1 dwarf (or 1:1 ratio of tall to dwarf) [2]
- Marking note: Cross Tt × tt produces offspring genotypes Tt and tt in 1:1 ratio. Tt = tall, tt = dwarf. Award 2 marks for correct ratio; 1 mark if ratio is correct but phenotypes are swapped.
Section C: Structured Questions (15 marks)
11. (a) A: Phosphate group / Phosphate [1] B: Deoxyribose sugar / Sugar [1] C: Nitrogenous base / Base [1] D: Hydrogen bond [1]
- Marking note: Accept "phosphate backbone" for A. Accept "pentose sugar" for B. Accept specific base names (adenine, thymine, guanine, cytosine) for C. Accept "H-bond" for D.
(b) DNA replicates by semi-conservative replication. The double helix unwinds and the two strands separate by breaking the hydrogen bonds between complementary bases. Each original strand serves as a template for the synthesis of a new complementary strand. Free nucleotides pair with exposed bases according to complementary base pairing rules (A with T, C with G). This produces two identical DNA molecules, each containing one original strand and one newly synthesised strand. [3]
- Marking note: Award 1 mark for mentioning unwinding/strand separation, 1 mark for template mechanism/complementary base pairing, and 1 mark for semi-conservative nature (one old + one new strand).
12. (a) (i) hh [1] (ii) Hh [1]
- Marking note: Affected female must be homozygous recessive (hh). Her husband is unaffected but they have an affected child, so he must be a carrier (heterozygous Hh).
(b) Both parents in Generation I are heterozygous (Hh). They are carriers of the sickle cell allele but do not have the disease because the normal allele (H) is dominant over the sickle cell allele (h). When two heterozygous individuals reproduce, there is a 25% chance that their child will inherit the recessive allele from both parents (hh) and thus be affected by sickle cell anaemia. [3]
- Marking note: Award 1 mark for stating both parents are heterozygous/carriers (Hh), 1 mark for explaining that the normal allele is dominant so carriers are unaffected, and 1 mark for explaining the 25% probability of an affected child from two heterozygous parents.
(c) Genetic diagram:
Parental genotypes: Hh × Hh
| H | h | |
|---|---|---|
| H | HH | Hh |
| h | Hh | hh |
Offspring genotypes: HH, Hh, Hh, hh Probability of sickle cell anaemia (hh) = 1/4 or 25%
Probability: 1/4 or 25% [3]
- Marking note: Award 1 mark for correct Punnett square setup, 1 mark for correct genotypes in the square, and 1 mark for correct probability (1/4 or 25%). Accept 0.25.
13. (a) (i) RR [1] (ii) rr [1] (iii) Rr [1]
- Marking note: Pure-breeding red = homozygous dominant (RR); pure-breeding white = homozygous recessive (rr); F₁ offspring from RR × rr = all heterozygous (Rr).
(b) The allele for red flower colour (R) is dominant over the allele for white flower colour (r). All F₁ offspring inherited one dominant allele (R) from the red parent and one recessive allele (r) from the white parent, giving them the genotype Rr. Since the dominant allele masks the expression of the recessive allele, all F₁ plants display the red flower phenotype. [2]
- Marking note: Award 1 mark for stating that red is dominant over white, and 1 mark for explaining that the dominant allele masks the recessive allele in heterozygous condition.
(c) 3 red : 1 white (or 3:1) [1]
- Marking note: F₁ self-pollination (Rr × Rr) produces offspring in expected genotypic ratio 1 RR : 2 Rr : 1 rr, giving phenotypic ratio 3 red : 1 white.
(d) The observed results (452 red : 148 white) are consistent with the expected Mendelian ratio of 3:1.
Expected numbers for 600 plants:
- Red: 3/4 × 600 = 450
- White: 1/4 × 600 = 150
The observed numbers (452 red, 148 white) are very close to the expected numbers (450 red, 150 white). The small difference can be attributed to chance variation in fertilisation and sampling. The ratio of 452:148 simplifies to approximately 3.05:1, which is very close to the expected 3:1 ratio. Therefore, the results support Mendelian inheritance of flower colour in this plant species. [4]
- Marking note: Award 1 mark for stating the results are consistent with the 3:1 ratio, 1 mark for calculating expected numbers (450 and 150), 1 mark for comparing observed and expected values, and 1 mark for explaining the small difference as due to chance/random fertilisation.
14. A dominant allele is one that is always expressed in the phenotype, even when only one copy is present (heterozygous condition). A recessive allele is only expressed in the phenotype when two copies are present (homozygous condition); it is masked by the dominant allele in a heterozygous individual. [2]
- Marking note: Award 1 mark for correct description of dominant allele (expressed in heterozygous condition) and 1 mark for correct description of recessive allele (expressed only in homozygous condition / masked by dominant allele).
15. Punnett square:
| IA | i | |
|---|---|---|
| IB | IAIB | IBi |
| i | IAi | ii |
Possible blood types and probabilities:
- Blood type AB (IAIB): 1/4 or 25%
- Blood type A (IAi): 1/4 or 25%
- Blood type B (IBi): 1/4 or 25%
- Blood type O (ii): 1/4 or 25%
Probability of each blood type: AB: 25%, A: 25%, B: 25%, O: 25% [3]
- Marking note: Award 1 mark for correct Punnett square setup, 1 mark for correct genotypes in the square, and 1 mark for correct probabilities for all four blood types.
Section D: Data-Based and Application Questions (10 marks)
16. (a) Parent genotypes: Rr (heterozygous round) and rr (wrinkled) [1]
- Marking note: Must state both genotypes correctly.
(b) Punnett square:
| R | r | |
|---|---|---|
| r | Rr | rr |
| r | Rr | rr |
| [2] |
- Marking note: Award 1 mark for correct gametes on axes, 1 mark for correct offspring genotypes in the square.
(c) Phenotypic ratio: 1 round : 1 wrinkled (or 1:1) [1]
- Marking note: Rr = round, rr = wrinkled.
17. A gene mutation is a change in the sequence of DNA bases. This can alter the sequence of codons in the mRNA transcribed from the gene. During translation, the altered codon sequence may code for a different amino acid, leading to a change in the amino acid sequence of the polypeptide. This can change the folding and shape of the protein, potentially affecting its function. [3]
- Marking note: Award 1 mark for mentioning change in DNA base sequence, 1 mark for linking to change in mRNA/codons, and 1 mark for explaining change in amino acid sequence/protein structure/function.
18. The probability that the fourth child will be a boy is 1/2 or 50%. The sex of a child is determined by the sex chromosome inherited from the father (X or Y). Each pregnancy is an independent event, and the probability of inheriting a Y chromosome (resulting in a male) is always 1/2, regardless of the sexes of previous children. [2]
- Marking note: Award 1 mark for stating 1/2 or 50%, and 1 mark for explaining independent assortment/previous children do not affect probability.
19. Homozygous means having two identical alleles for a particular gene. Example: AA or aa. [1]
- Marking note: Award 1 mark for correct definition and example. Accept either AA or aa as the example.
20. Advantage: Allows early detection of genetic disorders, enabling early intervention, treatment, or informed reproductive choices. [1] Disadvantage: May lead to discrimination, psychological stress, or ethical concerns about termination of pregnancy. [1]
- Marking note: Award 1 mark for a valid advantage and 1 mark for a valid disadvantage. Accept other reasonable answers.
END OF ANSWER KEY