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S5 Biology: Genetics and Mendelian Inheritance (Paper 3 (challenge))

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.

Direct answers Separate all from at least one: multiply directly for all, and subtract the complement for at least one. In pedigrees remember a father passes only Y to his sons.

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)

  1. FoundationTwo tall Tt plants give 480 offspring. How many tall and how many short are expected?
  2. FoundationA test cross gives a 1 : 1 phenotypic ratio. What is the genotype of the unknown parent?
  3. FoundationAaBb x AaBb gives 1600 plants with independent assortment. How many show both dominant traits A_B_?
  4. IntermediateTwo Tt parents have three children. What is the probability that at least one shows the recessive trait?
  5. IntermediateA couple of AB (IA IB) and O (ii) have what probability of a type O child?
  6. IntermediateA colour-blind father XcY and a normal homozygous mother XCXC have daughters of what kind?
  7. ChallengeTwo carriers Tt x Tt have two children. What is the probability that both show the recessive trait?
  8. ChallengeWhat blood groups can the children of O (ii) and O (ii) parents have?
  9. ChallengeA plant self-crosses and its offspring show a 3 : 1 ratio. What is its genotype?
  10. 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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