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4.8.4Additional lessons

Sex Determination — IB Biology HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 71 of 101

📚 IB Biology HL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
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Part 1 of 5Human Sex Chromosomes

XX and XY
  • XX and XY Females have two X chromosomes and males have one X and one Y, so sex is decided by a whole chromosome pair.
SRY gene
  • SRY gene The Y chromosome carries the SRY gene, which produces a protein that triggers development of testes in the embryo.
Role of gametes
  • Role of gametes All eggs carry an X chromosome, but half of the sperm carry X and half carry Y, so the sperm decides the sex.
Probability
  • Probability Each fertilisation has a 50% chance of producing a boy and a 50% chance of producing a girl.
ParentGametes producedSex chromosomes
Mother (XX)All XOne type of egg
Father (XY)Half X, half YTwo types of sperm

Part 2 of 5Sex Linkage

Sex Linkage
Sex-linked gene
  • Sex-linked gene A gene located on a sex chromosome, so the sex of the individual affects how the alleles are inherited.
X versus Y
  • X versus Y The X chromosome is much larger and carries many genes with no matching locus on the Y chromosome.
Males
  • Males Males have only one X, so a single recessive allele on it is expressed, because there is no second allele to mask it.
Females
  • Females Females need two recessive alleles to show the condition, and heterozygous females are carriers.
Examples
  • Examples Haemophilia and red-green colour blindness are caused by recessive alleles on the X chromosome.

Part 3 of 5Writing a Sex-Linked Cross

Alleles are written as superscripts on the X chromosome.

1
Use Xᴴ for the normal clotting allele and Xʰ for the haemophilia allele, and write the Y chromosome without any allele.
2
Write the parental genotypes, for example a carrier mother XᴴXʰ and a father with normal clotting XᴴY.
3
List the gametes: the mother makes Xᴴ and Xʰ eggs, the father makes Xᴴ and Y sperm.
4
Complete the Punnett grid and read off both genotype and sex of each child.
5
State the answer separately for sons and daughters, because the question may ask about one sex only.

Part 4 of 5Worked Example

Question

  • A carrier woman (XᴴXʰ) and a man with normal blood clotting (XᴴY) have a child.
  • Find the probability that the child has haemophilia, and the probability among sons.

  1. Gametesmother Xᴴ or Xʰ, father Xᴴ or Y.
  2. OffspringXᴴXᴴ, XᴴY, XᴴXʰ, XʰY, each with probability 1/4.
  3. Only XʰY has haemophilia, so the probability for any child is 1/4 = 25%.
  4. Among sons (XᴴY and XʰY) one of two is affected, so the probability is 1/2.

Answer

  • Any child: 25%;
    • a son: 50%;
    • no daughter will have haemophilia, but half will be carriers.

Part 5 of 5Exam Checklist & Tips

4.8.4 – Key Points for the Exam

  • Autosome Any chromosome that is not a sex chromosome.
  • SRY The gene on the Y chromosome that initiates male development.
  • Sex linkage Inheritance of a gene carried on a sex chromosome.
  • Carrier A heterozygous female who carries a recessive X-linked allele without showing the condition.

Common Mistakes

  • Writing sex-linked alleles on the Y chromosome when the gene is only on the X.
  • Forgetting that a father passes his X to every daughter and his Y to every son.
  • Stating that the mother's egg decides sex; it is the sperm that carries X or Y.
  • Giving ratios for all children when the question asks only about sons or daughters.

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