FORM FOUR BIOLOGY EXAM – TOPIC: GENETICS
Time: 3 Hours
Instructions:
- Answer all questions in Sections A and B.
- Answer two (2) questions only from Section C.
- All diagrams must be neat and properly labeled.
SECTION A (16 Marks)
1. Multiple Choice Questions (10 Marks)
Choose the most correct answer and write its letter:
- The physical expression of a gene is known as:
a) Genotype
b) Phenotype
c) Allele
d) Chromosome - In incomplete dominance, the heterozygous phenotype is:
a) Identical to the dominant phenotype
b) A blend of both alleles
c) Identical to the recessive phenotype
d) Not expressed - Which of the following is an example of codominance?
a) Red and white flowers producing pink flowers
b) Blood type AB
c) Tall and short plants producing medium height
d) None of the above - Haemophilia is a:
a) Dominant autosomal trait
b) Recessive autosomal trait
c) Dominant sex-linked trait
d) Recessive sex-linked trait - Albinism is caused by:
a) A dominant allele
b) A recessive allele
c) A codominant allele
d) An incomplete dominant allele - Epistasis refers to:
a) A gene that masks the expression of another gene
b) The blending of two traits
c) Genes located on sex chromosomes
d) The expression of recessive traits only - In humans, the gene for color blindness is located on:
a) Autosomes
b) X chromosome
c) Y chromosome
d) Mitochondrial DNA - The genotype IAIB represents blood type:
a) A
b) B
c) AB
d) O - A person with genotype XhXh will:
a) Be a carrier for hemophilia
b) Be normal
c) Have hemophilia
d) Be male - Which of the following crosses will result in a 1:2:1 genotypic ratio?
a) AA × aa
b) Aa × Aa
c) Aa × aa
d) AA × Aa
2. Matching Items (6 Marks)
Match the items in List A with the correct responses from List B:
| List A | List B |
|---|---|
| (i) Incomplete dominance | ___ |
| (ii) Codominance | ___ |
| (iii) Epistasis | ___ |
| (iv) Sex-linked trait | ___ |
| (v) Albinism | ___ |
| (vi) Hemophilia | ___ |
List B:
- Both alleles are fully expressed in heterozygotes
- A gene on the X chromosome affecting males more
- A recessive disorder causing lack of pigment
- One gene masks the effect of another gene
- A blending of traits in heterozygotes
- A recessive sex-linked bleeding disorder
SECTION B (54 Marks)
Answer all questions.
3.
(a) In humans, the allele for normal skin pigmentation (A) is dominant over albinism (a). A man and a woman, both heterozygous for this trait, plan to have a child. Use a Punnett square to determine the genotypic and phenotypic ratios of their offspring. (5 marks)
(b) What is the probability that their child will have albinism? (4 marks)
4.
(a) In snapdragons, flower color exhibits incomplete dominance. Red flowers (RR) crossed with white flowers (WW) produce pink flowers (RW). If a pink-flowered plant is crossed with a white-flowered plant, use a Punnett square to determine the genotypic and phenotypic ratios of the offspring. (5 marks)
(b) Explain how incomplete dominance differs from complete dominance. (4 marks)
5.
(a) In cattle, coat color shows codominance. Red coat color is represented by CRCR, white by CWCW, and roan (a mix of red and white hairs) by CRCW. If two roan cattle are crossed, use a Punnett square to determine the genotypic and phenotypic ratios of their offspring. (5 marks)
(b) What is the probability of producing a calf with a red coat? (4 marks)
6.
(a) In humans, the allele for normal blood clotting (H) is dominant over hemophilia (h), a sex-linked recessive disorder. A woman who is a carrier (XHXh) marries a man with normal blood clotting (XHY). Use a Punnett square to determine the genotypic and phenotypic ratios of their children. (5 marks)
(b) What is the probability that their son will have hemophilia? (4 marks)
7.
(a) In fruit flies, eye color is a sex-linked trait. Red eyes (Xᴿ) are dominant over white eyes (Xʷ). A red-eyed female heterozygote (XᴿXʷ) is crossed with a white-eyed male (XʷY). Use a genetic cross to determine the genotypic and phenotypic ratios of the offspring. (5 marks)
(b) What percentage of female offspring will have white eyes? (4 marks)
8.
(a) In humans, the allele for normal skin pigment (A) is epistatic to the allele for melanin production (B/b). If a person has aa genotype, they will be albino regardless of the B allele. If two individuals with AaBb genotypes are crossed, show the phenotypic ratio of their offspring. (5 marks)
(b) Explain the role of the epistatic gene in this cross. (4 marks)
SECTION C (30 Marks)
Answer any TWO questions. Each carries 15 marks.
9.
Describe the structure and function of DNA and RNA. Explain how mutations in genetic material can lead to genetic disorders in humans.
10.
Discuss the differences between autosomal dominant, autosomal recessive, and sex-linked inheritance. Use suitable genetic disorders as examples to support your answer.
11.
With the aid of genetic diagrams, explain how codominance and incomplete dominance differ. Use real-life examples to show their genetic significance in inheritance.
FORM FOUR BIOLOGY EXAM ANSWERS – TOPIC: GENETICS
SECTION A (16 Marks)
1. Multiple Choice Questions (10 Marks)
(i) b) Phenotype
(ii) b) A blend of both alleles
(iii) b) Blood type AB
(iv) d) Recessive sex-linked trait
(v) b) A recessive allele
(vi) a) A gene that masks the expression of another gene
(vii) b) X chromosome
(viii) c) AB
(ix) c) Have hemophilia
(x) b) Aa × Aa
2. Matching Items (6 Marks)
(i) Incomplete dominance - e) A blending of traits in heterozygotes
(ii) Codominance - a) Both alleles are fully expressed in heterozygotes
(iii) Epistasis - d) One gene masks the effect of another gene
(iv) Sex-linked trait - b) A gene on the X chromosome affecting males more
(v) Albinism - c) A recessive disorder causing lack of pigment
(vi) Hemophilia - f) A recessive sex-linked bleeding disorder
SECTION B (54 Marks)
3. (a) Punnett square for albinism cross (5 marks)
| A | a | |
|---|---|---|
| A | AA (normal) | Aa (normal) |
| a | Aa (normal) | aa (albino) |
Genotypic ratio: 1 AA : 2 Aa : 1 aa
Phenotypic ratio: 3 normal : 1 albino
(b) Probability of albinism (4 marks)
Probability = 1/4 or 25%
4. (a) Punnett square for incomplete dominance (5 marks)
| R | W | |
|---|---|---|
| W | RW (pink) | WW (white) |
| W | RW (pink) | WW (white) |
Genotypic ratio: 2 RW : 2 WW
Phenotypic ratio: 2 pink : 2 white (or 1:1)
(b) Incomplete vs complete dominance (4 marks)
Incomplete dominance: Heterozygote shows intermediate phenotype (blending). Complete dominance: Heterozygote shows same phenotype as homozygous dominant.
5. (a) Punnett square for codominance (5 marks)
| CR | CW | |
|---|---|---|
| CR | CRCR (red) | CRCW (roan) |
| CW | CRCW (roan) | CWCW (white) |
Genotypic ratio: 1 CRCR : 2 CRCW : 1 CWCW
Phenotypic ratio: 1 red : 2 roan : 1 white
(b) Probability of red calf (4 marks)
Probability = 1/4 or 25%
6. (a) Punnett square for hemophilia (5 marks)
| XH | Y | |
|---|---|---|
| XH | XHXH (normal female) | XHY (normal male) |
| Xh | XHXh (carrier female) | XhY (hemophiliac male) |
Genotypic ratio: 1 XHXH : 1 XHXh : 1 XHY : 1 XhY
Phenotypic ratio: 1 normal female : 1 carrier female : 1 normal male : 1 hemophiliac male
(b) Probability of son with hemophilia (4 marks)
Probability = 1/2 or 50%
7. (a) Genetic cross for eye color (5 marks)
| Xʷ | Y | |
|---|---|---|
| Xᴿ | XᴿXʷ (red-eyed female) | XᴿY (red-eyed male) |
| Xʷ | XʷXʷ (white-eyed female) | XʷY (white-eyed male) |
Genotypic ratio: 1 XᴿXʷ : 1 XʷXʷ : 1 XᴿY : 1 XʷY
Phenotypic ratio: 1 red-eyed female : 1 white-eyed female : 1 red-eyed male : 1 white-eyed male
(b) Percentage of white-eyed females (4 marks)
Percentage = 50% (1 out of 2 female genotypes)
8. (a) Phenotypic ratio for epistasis (5 marks)
Phenotypic ratio: 9 normal pigment : 3 albino : 4 (depends on specific epistatic interaction)
(b) Role of epistatic gene (4 marks)
The epistatic gene (A/a) controls whether melanin can be produced at all. Even if the B gene for melanin production is present, if the aa genotype exists, no pigment will be produced (albinism).
SECTION C (30 Marks)
9. DNA/RNA structure and mutations (15 marks)
DNA structure: Double helix with sugar-phosphate backbone, nitrogenous bases (A,T,C,G), complementary base pairing.
RNA structure: Single-stranded, contains uracil instead of thymine, various types (mRNA, tRNA, rRNA).
Function: DNA stores genetic information, RNA helps in protein synthesis.
Mutations: Changes in DNA sequence can lead to altered proteins. Examples: point mutations (sickle cell anemia), frameshift mutations, chromosomal abnormalities (Down syndrome). Mutations can be inherited or acquired.
10. Inheritance patterns (15 marks)
Autosomal dominant: Only one copy needed to express trait. Example: Huntington's disease. Affects both sexes equally.
Autosomal recessive: Two copies needed to express trait. Example: Cystic fibrosis. Carriers unaffected.
Sex-linked: Genes on X chromosome. Males more affected (XY). Examples: Hemophilia, color blindness. Females need two copies to show recessive trait.
Patterns differ in inheritance probability, gender distribution, and carrier status.
11. Codominance vs incomplete dominance (15 marks)
Codominance: Both alleles fully expressed in heterozygote (e.g., AB blood type, roan cattle). No blending - both phenotypes visible.
Incomplete dominance: Intermediate phenotype in heterozygote (e.g., pink snapdragons from red and white parents). Blending occurs.
Genetic significance: Shows that dominance relationships between alleles can vary. Important in understanding complex inheritance patterns beyond simple Mendelian genetics.
Both demonstrate that not all traits follow simple dominant/recessive patterns, increasing genetic diversity in populations.
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