Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

4.8.11Additional lessons

Identifying Recombinants — IB Biology HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 78 of 101

📚 IB Biology HL Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read

Part 1 of 5Crossing Over and Recombinants

4.8.11 Identifying Recombinants
Crossing over
  • In prophase I of meiosis, non-sister chromatids of a homologous pair exchange segments at points called chiasmata.
Recombinant chromatid
  • A chromatid with a new combination of alleles, such as (Tg) or (tG), produced after segments have been exchanged.
Recombinant offspring
  • Offspring with a phenotype combination that differs from both parents, produced from recombinant gametes.
Parental type
  • Offspring with the same combination of characteristics as the parents, produced from gametes without a crossover between the genes.
Source of variation
  • Crossing over, together with independent assortment, increases genetic variation in a population.

Part 2 of 5Recombinant Frequency

Recombinant Frequency
Definition
  • The percentage of offspring that are recombinants, found as the number of recombinants divided by the total offspring, × 100.
Distance effect
  • Genes that are far apart on a chromosome are separated more often by a chiasma, so their recombinant frequency is higher.
Close genes
  • Genes very close together have a low recombinant frequency, because a chiasma rarely forms between them.
Upper limit
  • The frequency cannot exceed 50%; at 50% the genes behave as if they were unlinked and gave 1 : 1 : 1 : 1.
GenesRecombinant frequencyBehaviour
Very closeLowAlmost always inherited together
Far apartUp to 50%Assort almost independently

Part 3 of 5Recognising Non-Mendelian Ratios

1
Predict the ratio expected if the genes were unlinked, such as 9 : 3 : 3 : 1 in an F₂ or 1 : 1 : 1 : 1 in a test cross.
2
Compare the real numbers: linked genes give far more parental types than expected and fewer recombinants.
3
Pick out the two smallest classes in a test cross, which are normally the recombinant phenotypes.
4
Calculate the recombinant frequency from the totals, and use a chi-squared test if you need to check that the ratio differs significantly from the unlinked ratio.
5
Explain the result using linkage and crossing over, not sex linkage.

Part 4 of 5Worked Example

Question

  • A test cross of a double heterozygote gives 1000 offspring: 442 with phenotype AB, 438 with ab, 62 with Ab and 58 with aB.
  • Identify the recombinants and calculate the recombinant frequency.

  1. The two largest classes (AB and ab) are the parental types, since they match the original chromosomes.
  2. The two smallest classes (Ab and aB) are the recombinants.
  3. Number of recombinants = 62 + 58 = 120.
  4. Recombinant frequency = 120 ÷ 1000 × 100 = 12%.

Answer

  • The recombinants are Ab and aB;
    • the recombinant frequency is 12%, which shows the genes are linked but not very close.

Part 5 of 5Exam Checklist & Tips

4.8.11 – Key Points for the Exam

  • Recombinant An offspring (or gamete) with a combination of alleles different from either parent.
  • Chiasma The point where non-sister chromatids cross over during prophase I.
  • Recombinant frequency The percentage of offspring that are recombinants.
  • Test cross A cross with a homozygous recessive individual used to reveal gamete types.

Common Mistakes

  • Counting the parental classes as recombinants because they are the largest.
  • Dividing the number of recombinants by the number of parental offspring instead of by the total.
  • Stating that crossing over happens between sister chromatids.
  • Saying that linked genes can never be separated.

Last updated:

\n