S5 Biology: Genetics and Mendelian Inheritance (Paper 3 (challenge))
- 2026-11-14
S5 Biology: Genetics and Mendelian Inheritance (Paper 3 (challenge))
This paper extends single-gene ratios to independent assortment of two genes, focuses on the opposite probability questions of at least one and all, and checks the direction of X-linked transmission.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Biology: Genetics and Mendelian Inheritance |
| Level | S3 |
02 Questions (10 items)
- FoundationTwo tall Tt plants give 480 offspring. How many tall and how many short are expected?
- FoundationA test cross gives a 1 : 1 phenotypic ratio. What is the genotype of the unknown parent?
- FoundationAaBb x AaBb gives 1600 plants with independent assortment. How many show both dominant traits A_B_?
- IntermediateTwo Tt parents have three children. What is the probability that at least one shows the recessive trait?
- IntermediateA couple of AB (IA IB) and O (ii) have what probability of a type O child?
- IntermediateA colour-blind father XcY and a normal homozygous mother XCXC have daughters of what kind?
- ChallengeTwo carriers Tt x Tt have two children. What is the probability that both show the recessive trait?
- ChallengeWhat blood groups can the children of O (ii) and O (ii) parents have?
- ChallengeA plant self-crosses and its offspring show a 3 : 1 ratio. What is its genotype?
- Error redoError clinic: a student claims the same phenotype means the same genotype. What is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 360 tall and 120 short. | In a 3 : 1 ratio, 480 times 3 over 4 is 360 and the rest is 120. |
| 2 | A heterozygote such as Tt. | Only a heterozygote produces two kinds of alleles, giving 1 : 1. |
| 3 | 900 plants. | A_B_ takes nine sixteenths of a dihybrid F2, and 1600 times 9 over 16 is 900. |
| 4 | Thirty-seven sixty-fourths, about 57.8 percent. | All three dominant has probability 27/64, so at least one recessive is 64 minus 27, giving 37. |
| 5 | Zero percent, a type O child is impossible. | The AB parent carries no i allele, so children can only be A or B. |
| 6 | All daughters are phenotypically normal but all are carriers XCXc. | The father passes Xc to every daughter while the mother supplies XC. |
| 7 | One sixteenth, that is 1/16. | Each child has probability one quarter, and 4 x 4 gives 16. |
| 8 | All are type O, that is 100 percent. | Both parents carry only i, so only i can be passed on. |
| 9 | A heterozygote such as Tt. | A homozygote does not segregate; only a heterozygote self-cross gives 3 : 1. |
| 10 | Not necessarily, since TT and Tt are both tall. | The dominant allele masks the difference in a heterozygote, so a test cross or a DNA test is needed. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Sort reproduction questions by place and order first: the testes make sperm, the ovaries make eggs, fertilisation happens in the oviduct and implantation in the uterus, while the placenta and umbilical cord carry nutrients and oxygen; for cycle questions draw the days on one timeline and remember the fertile window runs about three days before and one day after ovulation.Tutor Registration →
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