4.8.11Additional lessons
Identifying Recombinants — IB Biology HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 78 of 101
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
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5Crossing Over and Recombinants
Crossing over
- Crossing over In prophase I of meiosis, non-sister chromatids of a homologous pair exchange segments at points called chiasmata.
Recombinant chromatid
- Recombinant chromatid A chromatid with a new combination of alleles, such as (Tg) or (tG), produced after segments have been exchanged.
Recombinant offspring
- Recombinant offspring Offspring with a phenotype combination that differs from both parents, produced from recombinant gametes.
Parental type
- Parental type Offspring with the same combination of characteristics as the parents, produced from gametes without a crossover between the genes.
Source of variation
- Source of variation Crossing over, together with independent assortment, increases genetic variation in a population.
Part 2 of 5Recombinant Frequency
Recombinant Frequency
Definition
- Definition The percentage of offspring that are recombinants, found as the number of recombinants divided by the total offspring, × 100.
Distance effect
- 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
- Close genes Genes very close together have a low recombinant frequency, because a chiasma rarely forms between them.
Upper limit
- Upper limit The frequency cannot exceed 50%;
- at 50% the genes behave as if they were unlinked and gave 1 : 1 : 1 : 1.
| Genes | Recombinant frequency | Behaviour |
|---|---|---|
| Very close | Low | Almost always inherited together |
| Far apart | Up 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.
- The two largest classes (AB and ab) are the parental types, since they match the original chromosomes.
- The two smallest classes (Ab and aB) are the recombinants.
- Number of recombinants = 62 + 58 = 120.
- 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.
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