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Secondary 4 Combined Science Biology Genetics Inheritance Quiz
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Questions
Secondary 4 Combined Science Biology Quiz - Genetics Inheritance
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 40
Duration: 45 minutes Total Marks: 40
Instructions:
- This quiz contains 20 questions on Genetics & Inheritance.
- Answer ALL questions in the spaces provided.
- Show all working for genetic diagrams and calculations.
- The number of marks is shown in brackets [ ] at the end of each question or part question.
Section A: Multiple Choice Questions (5 marks)
Circle the correct answer for each question.
1. Which of the following best describes a gene?
- A) A length of DNA that codes for a specific protein
- B) A structure made of protein that carries genetic information
- C) The physical appearance of an organism
- D) An alternative form of a characteristic
[1]
2. In humans, the allele for brown eyes (B) is dominant over the allele for blue eyes (b). What is the genotype of a person who is homozygous recessive for eye colour?
- A) BB
- B) Bb
- C) bB
- D) bb
[1]
3. A diploid human cell contains 46 chromosomes. How many chromosomes are present in a human sperm cell?
- A) 46
- B) 23
- C) 92
- D) 22
[1]
4. Which process produces gametes with half the number of chromosomes?
- A) Mitosis
- B) Binary fission
- C) Meiosis
- D) Fertilisation
[1]
5. A man with blood group AB marries a woman with blood group O. Which blood groups are possible in their children?
- A) A and B only
- B) AB and O only
- C) A, B, and AB only
- D) A, B, AB, and O
[1]
Section B: Structured Questions (20 marks)
Answer all questions in the spaces provided.
6. Distinguish between the terms genotype and phenotype.
Genotype: _________________________________________________________________
Phenotype: ________________________________________________________________ [2]
7. Explain why an individual can have two alleles for a gene but produce only one type of gamete with respect to that gene.
[2]
8. In pea plants, the allele for round seeds (R) is dominant over the allele for wrinkled seeds (r).
(a) State the genotype of a pea plant that is heterozygous for seed shape.
[1]
(b) Two heterozygous round-seeded pea plants are crossed. Using a genetic diagram, determine the expected phenotypic ratio of the offspring.
Parental phenotypes: ________________________________________________________
Parental genotypes: ________________________________________________________
Gametes: _________________________________________________________________
Offspring genotypes: _______________________________________________________
Offspring phenotypes: ______________________________________________________
Phenotypic ratio: __________________________________________________________ [4]
9. A student states: "Dominant alleles are always more common in a population than recessive alleles." Explain why this statement is incorrect, using a named example.
[2]
10. The diagram below shows a pedigree chart for a family with a history of a genetic disorder. Affected individuals are shaded.
I 1 ─── 2
│
II 1 ─── 2 3 ─── 4
│ │
III 1 2 3 4 5
█ █
(a) Using evidence from the pedigree, determine whether the disorder is caused by a dominant or recessive allele. Explain your reasoning.
[2]
(b) Individual II-2 is pregnant with her fourth child. What is the probability that this child will be affected by the disorder? Explain your answer.
[2]
Section C: Structured Questions (continued) (10 marks)
Answer all questions in the spaces provided.
11. Explain how meiosis contributes to genetic variation in offspring.
[3]
12. In fruit flies, the allele for red eyes (R) is dominant over the allele for white eyes (r). The gene for eye colour is located on the X chromosome.
(a) Explain why male fruit flies are more likely to express the white-eye phenotype than female fruit flies.
[2]
13. A geneticist studied the inheritance of coat colour in rabbits. The allele for black coat (B) is dominant over the allele for white coat (b). The geneticist carried out three crosses and recorded the results in the table below.
| Cross | Parental Phenotypes | Offspring Phenotypes |
|---|---|---|
| 1 | Black × White | All black |
| 2 | Black × Black | 3 black : 1 white |
| 3 | Black × White | 1 black : 1 white |
(a) For Cross 1, deduce the genotypes of both parents. Explain your reasoning.
[2]
(b) For Cross 3, explain why the offspring ratio is 1:1 rather than all black.
[2]
14. Cystic fibrosis is a genetic disorder caused by a recessive allele. A man and a woman, both of whom are carriers of the cystic fibrosis allele, are planning to have children.
(a) Using a genetic diagram, determine the probability that their first child will have cystic fibrosis.
Parental genotypes: ________________________________________________________
Gametes: _________________________________________________________________
Offspring genotypes: _______________________________________________________
Offspring phenotypes: ______________________________________________________
Probability of child with cystic fibrosis: ______________________________________ [3]
Section D: Data-Based and Extended Response Questions (5 marks)
Answer all questions in the spaces provided.
15. A black rabbit from Cross 2 in Question 13 is selected at random. Describe how you could determine whether this rabbit is homozygous or heterozygous for coat colour.
[3]
16. The couple in Question 14 has their first child born without cystic fibrosis. Explain why the probability that their second child will have cystic fibrosis remains the same as for the first child.
[2]
17. Sickle cell anaemia is caused by a recessive allele. In certain regions of Africa, the frequency of the sickle cell allele is relatively high despite the disorder being harmful. Heterozygous individuals have some resistance to malaria.
(a) Explain why the sickle cell allele persists in these populations despite its harmful effects in homozygous individuals.
[2]
(b) State the term used to describe a situation where a heterozygous individual has an advantage over both homozygous genotypes.
[1]
18. Explain the difference between continuous and discontinuous variation, giving one example of each in humans.
[2]
19. A man with blood group A claims he is the father of a child with blood group O. The mother has blood group B. Using a genetic diagram, determine whether the man could be the father. Explain your reasoning.
[3]
20. Describe how a mutation in a gene can lead to the production of a non-functional protein.
[2]
END OF QUIZ
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Answers
Secondary 4 Combined Science Biology Quiz - Genetics Inheritance
ANSWER KEY AND MARKING SCHEME
Total Marks: 40
Section A: Multiple Choice Questions (5 marks)
1. A) A length of DNA that codes for a specific protein [1]
- Explanation: A gene is a segment of DNA that carries the code for making a specific polypeptide or protein. Option B describes a chromosome, C describes phenotype, and D describes an allele.
2. D) bb [1]
- Explanation: Homozygous means both alleles are the same. Recessive means the allele is only expressed when homozygous. Therefore, homozygous recessive for eye colour is bb.
3. B) 23 [1]
- Explanation: Sperm cells are gametes produced by meiosis, which halves the chromosome number. A diploid cell (46 chromosomes) produces haploid gametes (23 chromosomes).
4. C) Meiosis [1]
- Explanation: Meiosis is the type of cell division that produces gametes with half the number of chromosomes (haploid). Mitosis produces identical diploid cells.
5. A) A and B only [1]
- Explanation: A person with blood group AB has genotype I^A I^B. A person with blood group O has genotype I^O I^O. Possible offspring genotypes: I^A I^O (blood group A) and I^B I^O (blood group B).
Section B: Structured Questions (20 marks)
6.
- Genotype: The genetic makeup of an organism; the combination of alleles an organism possesses for a particular characteristic. [1]
- Phenotype: The observable physical or biochemical characteristics of an organism, resulting from the interaction of its genotype with the environment. [1]
- Marking note: Accept equivalent definitions. Must convey that genotype is the genetic constitution and phenotype is the expressed trait.
7. An individual has two alleles for each gene (one on each homologous chromosome). During meiosis, homologous chromosomes separate so that each gamete receives only one allele of each gene. Therefore, although the individual possesses two alleles, each gamete contains only one. [2]
- Marking notes: 1 mark for stating homologous chromosomes separate during meiosis; 1 mark for explaining that each gamete receives only one allele. Accept reference to Mendel's Law of Segregation.
8. (a) Rr [1]
- Marking note: Must use correct allele symbols as defined in the question.
(b)
- Parental phenotypes: Round seeds × Round seeds [0.5]
- Parental genotypes: Rr × Rr [0.5]
- Gametes: R, r and R, r [0.5]
- Offspring genotypes: RR, Rr, Rr, rr [1]
- Offspring phenotypes: Round, Round, Round, Wrinkled [0.5]
- Phenotypic ratio: 3 round : 1 wrinkled [1]
- Marking notes: Award marks for correct genetic diagram format. Must show all steps. Accept Punnett square or branching diagram. Deduct 0.5 marks if ratio is not simplified or if genotype and phenotype ratios are confused.
9. The statement is incorrect because dominance refers to which allele is expressed in a heterozygous individual, not how common the allele is in a population. For example, the allele for polydactyly (having extra fingers/toes) is dominant, but it is very rare in human populations. Conversely, the allele for type O blood is recessive but is very common in many populations. [2]
- Marking notes: 1 mark for explaining that dominance does not determine allele frequency; 1 mark for a valid named example. Accept other valid examples (e.g., Huntington's disease is dominant but rare; blue eyes are recessive but common in some populations).
10. (a) The disorder is caused by a recessive allele. Evidence: 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 to pass on the dominant allele. Since both parents are unaffected but have an affected child, they must be heterozygous carriers of a recessive allele. [2]
- Marking notes: 1 mark for correct determination (recessive); 1 mark for valid explanation using evidence from the pedigree. Must reference specific individuals.
(b) Probability = 0. Individual II-2 is unaffected but must be a carrier (heterozygous) because she has an affected sibling (II-3) and her parents are carriers. Her partner (II-1) is also unaffected and unrelated, so he is assumed to be homozygous normal. Cross: Carrier female (Aa) × Normal male (AA). All offspring will be unaffected; none will have the disorder. [2]
- Marking notes: 1 mark for identifying II-2 as a carrier; 1 mark for correct probability with explanation. Accept alternative reasoning if student assumes II-1 could be a carrier (then probability would be 1/4 × 2/3 = 1/6, but this is beyond Combined Science scope). The simpler answer (0) is acceptable given the pedigree information.
Section C: Structured Questions (continued) (10 marks)
11. Meiosis contributes to genetic variation through:
- Independent assortment: During meiosis I, homologous chromosomes line up randomly at the equator. The orientation of each pair is independent of other pairs, resulting in different combinations of maternal and paternal chromosomes in gametes. [1.5]
- Crossing over: During prophase I, homologous chromosomes exchange segments of DNA at points called chiasmata. This creates new combinations of alleles on the same chromosome. [1.5]
- Marking notes: Award 1.5 marks for each mechanism clearly explained. Must mention both processes for full marks. Accept reference to random fertilisation as an additional source of variation (but this occurs after meiosis).
12. (a) Male fruit flies have only one X chromosome (XY), so they have only one allele for the eye colour gene. If they inherit the recessive white-eye allele (X^r) from their mother, they will express the white-eye phenotype because there is no dominant allele on the Y chromosome to mask it. Female fruit flies have two X chromosomes (XX), so they need two recessive alleles (X^r X^r) to express the white-eye phenotype. A heterozygous female (X^R X^r) will have red eyes because the dominant allele masks the recessive allele. [2]
- Marking notes: 1 mark for explaining that males have only one X chromosome/one allele; 1 mark for explaining that females need two recessive alleles to express the trait. Must reference X-linkage.
13. (a) Black parent: BB (homozygous dominant); White parent: bb (homozygous recessive). Reasoning: All offspring are black, which means the black parent must have passed on a dominant B allele to every offspring. If the black parent were heterozygous (Bb), approximately half the offspring would be white when crossed with a white rabbit (bb). Since all offspring are black, the black parent must be homozygous dominant (BB). [2]
- Marking notes: 1 mark for correct genotypes; 1 mark for valid reasoning. Must explain why the black parent cannot be heterozygous.
(b) In Cross 3, the black parent must be heterozygous (Bb) and the white parent is homozygous recessive (bb). The cross Bb × bb produces gametes B, b and b, b. Offspring genotypes: Bb (black) and bb (white) in a 1:1 ratio. This is a test cross. [2]
- Marking notes: 1 mark for identifying the black parent as heterozygous; 1 mark for explaining the 1:1 ratio using gametes and offspring genotypes.
14. (a)
- Parental genotypes: Cc × Cc (where C = normal allele, c = cystic fibrosis allele) [0.5]
- Gametes: C, c and C, c [0.5]
- Offspring genotypes: CC, Cc, Cc, cc [1]
- Offspring phenotypes: Normal, Normal (carrier), Normal (carrier), Cystic fibrosis [0.5]
- Probability of child with cystic fibrosis: 1/4 or 25% [0.5]
- Marking notes: Award marks for correct genetic diagram format. Must show all steps. Accept Punnett square.
Section D: Data-Based and Extended Response Questions (5 marks)
15. Perform a test cross: Cross the black rabbit of unknown genotype with a white rabbit (homozygous recessive, bb).
- If the black rabbit is homozygous (BB), all offspring will be black (Bb). [1]
- If the black rabbit is heterozygous (Bb), approximately half the offspring will be black (Bb) and half will be white (bb). [1]
- The appearance of any white offspring confirms the black rabbit is heterozygous. If a large number of offspring are all black, the rabbit is likely homozygous. [1]
- Marking notes: 1 mark for describing the test cross method; 1 mark for explaining the expected results for each genotype; 1 mark for explaining how to interpret the results. Accept reference to "breeding with a known homozygous recessive".
16. The probability remains the same (1/4 or 25%) because each fertilisation event is independent. The outcome of the first pregnancy does not affect the genetic makeup of the gametes produced for the second pregnancy. Each parent is still heterozygous (carrier), so the probability of passing on the recessive allele is still 1/2 for each parent, resulting in a 1/4 chance of having an affected child each time. [2]
- Marking notes: 1 mark for stating that each pregnancy is an independent event; 1 mark for explaining that the parental genotypes and gamete probabilities remain unchanged.
17. (a) In regions where malaria is prevalent, heterozygous individuals (carriers of the sickle cell allele) have a selective advantage because they have some resistance to malaria. This means they are more likely to survive and reproduce compared to individuals who are homozygous for the normal allele (who are susceptible to malaria) or homozygous for the sickle cell allele (who have sickle cell anaemia). As a result, the sickle cell allele is passed on to future generations and maintained in the population despite its harmful effects in homozygous individuals. This is an example of heterozygote advantage. [2]
- Marking notes: 1 mark for explaining the survival advantage of heterozygotes in malarial regions; 1 mark for linking this to the persistence of the allele through reproduction.
(b) Heterozygote advantage [1]
- Marking note: Accept "heterozygous advantage" or "balanced polymorphism".
18. Continuous variation is a type of variation where individuals show a range of phenotypes with no distinct categories (e.g., height, body mass). Discontinuous variation is where individuals fall into distinct, separate categories with no intermediates (e.g., blood groups, ability to roll tongue). [2]
- Marking notes: 1 mark for correct explanation of both types; 1 mark for one valid example of each in humans. Examples must be correct.
19. The man could be the father. The child has blood group O (genotype I^O I^O). The mother has blood group B, so her genotype could be I^B I^B or I^B I^O. For the child to inherit an I^O allele from each parent, the mother must be I^B I^O. The man with blood group A could have genotype I^A I^A or I^A I^O. If the man is I^A I^O, he can pass on the I^O allele. Cross: I^A I^O × I^B I^O. Possible offspring: I^A I^B (AB), I^A I^O (A), I^B I^O (B), I^O I^O (O). Therefore, a man with blood group A who is heterozygous can father a child with blood group O. [3]
- Marking notes: 1 mark for identifying the genotypes of child and mother; 1 mark for showing the man must be heterozygous I^A I^O; 1 mark for correct genetic diagram and conclusion. Deduct marks if student assumes the man cannot be the father without considering heterozygosity.
20. A mutation is a change in the DNA sequence of a gene. This can alter the sequence of amino acids in the protein it codes for. The change in amino acid sequence can affect the folding and three-dimensional shape of the protein. Since a protein's function depends on its specific shape, a change in shape can render the protein non-functional (e.g., an enzyme's active site may no longer bind to its substrate). [2]
- Marking notes: 1 mark for explaining that a mutation changes the DNA sequence leading to an altered amino acid sequence; 1 mark for explaining how this affects protein shape and function. Accept specific examples (e.g., sickle cell anaemia).
END OF ANSWER KEY