Numerical Problems/ISC/Class 12
Created by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
BioNotesBioNotesCreated by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
Online
Numerical Problems - Mendelian Crosses
Mastering genetics requires moving beyond simple monohybrid crosses and learning to apply probability laws, recognize linked genes, and analyze pedigrees logically. The following numerical problems are designed to test deep conceptual understanding and mathematical application in genetics.
In pea plants, tall stem (T) is dominant to dwarf (t), round seeds (R) is dominant to wrinkled (r), and yellow seeds (Y) is dominant to green (y). A cross is made between a fully heterozygous plant and a plant heterozygous for the first two traits but recessive for the third: TtRrYy × TtRryy.
Calculate the probability of obtaining an offspring that is:
Instead of drawing a massive 64-box Punnett square, we use the Product Rule of Probability (the Branching Method), treating each gene as an independent monohybrid cross.
Step 1: Break down the cross into three monohybrid crosses.
Step 2: Calculate specific probabilities.
Tall, wrinkled, yellow:
Genotype TtRrYy:
[!CAUTION] Common Pitfall / Trap: Students often try to draw a multi-gene Punnett square, which wastes time and leads to counting errors. Always check for independent assortment; if genes are on different chromosomes, the probability of simultaneous independent events is simply the product of their individual probabilities.
Below is a pedigree chart for an autosomal recessive trait (e.g., Albinism).
Pedigree Analysis Structure:
Individuals I-1 and I-2 are unaffected, but their first child (II-1) is affected. Their second child, II-2 (unaffected), marries II-3, who is a known carrier (heterozygous) for the trait. What is the probability that their child (III-1) will be affected?
Step 1: Determine the genotypes of the grandparents (I-1 and I-2). Since they are unaffected but have an affected child (aa), both I-1 and I-2 must be heterozygous carriers (Aa).
Step 2: Determine the probability that II-2 is a carrier. A cross between two heterozygotes (Aa × Aa) yields a genotypic ratio of 1 AA : 2 Aa : 1 aa. Since we already know by observation that II-2 is unaffected, we must eliminate the 'aa' possibility from our denominator.
Step 3: Calculate the probability of an affected child (III-1). For III-1 to be affected (aa), II-2 MUST be a carrier AND pass on the 'a' allele, AND II-3 MUST pass on the 'a' allele.
[!WARNING] Common Pitfall / Trap: The most frequent error is assuming that the probability of II-2 being a carrier is 1/2. Because II-2's phenotype is known (unaffected), the sample space shrinks from 4 to 3 (AA, Aa, aA). This is known as conditional probability.
Genes A and B are linked on the same chromosome. A plant heterozygous for both genes in the cis arrangement (AB/ab) is test-crossed with a homozygous recessive plant (ab/ab). The following 1,000 progeny are observed:
Calculate the recombination frequency and the map distance between genes A and B.
Linkage Phases:
Step 1: Identify parental and recombinant phenotypes. Because the heterozygous parent was in the cis arrangement (AB on one chromosome, ab on the other), the parental gametes are AB and ab.
Step 2: Calculate Recombination Frequency (RF).
Formula: RF = (Number of Recombinants / Total Offspring) × 100
Step 3: Determine Map Distance. 1% recombination frequency is equal to 1 map unit (m.u.) or 1 centiMorgan (cM).
[!TIP] Common Pitfall / Trap: If the heterozygous parent had been in the trans arrangement (Ab/aB), the dominant phenotypes for one trait combined with the recessive for the other (Aabb and aaBb) would be the highly populated parental types, and AaBb/aabb would be the rare recombinants. Always check the initial linkage phase!
In mice, coat color is determined by two unlinked genes. Gene C allows for pigment production (C_ = pigmented, cc = albino). Gene A determines the distribution of pigment (A_ = agouti/banded, aa = solid black).
A cross is performed between two dihybrid agouti mice (AaCc × AaCc). Determine the expected phenotypic ratio of their offspring.
This is a classic case of recessive epistasis, where the homozygous recessive condition at one locus (cc) masks the expression of alleles at a second locus (A/a).
Step 1: Determine the standard Mendelian dihybrid ratios.
Step 2: Apply the epistatic conditions.
Step 3: Combine identical phenotypes.
Final Phenotypic Ratio: 9 Agouti : 3 Black : 4 Albino
[!NOTE] Common Pitfall / Trap: Forgetting that epistasis alters standard 9:3:3:1 ratios. If a question gives you a 9:3:4, 12:3:1, or 9:7 phenotypic ratio from a dihybrid cross, you are immediately looking at an epistatic interaction, not a single gene or linked genes.
Human skin color is a quantitative trait controlled by multiple polygenes. Assume skin color is controlled by 3 independently assorting genes (A, B, C) with additive effects. The dominant alleles (A, B, C) add pigment, while recessive alleles (a, b, c) do not.
Two individuals with intermediate skin color, both with genotype AaBbCc, have children. What is the probability of them having a child with exactly 5 pigment-producing (dominant) alleles?
For quantitative inheritance involving multiple unlinked genes, phenotypes depend on the total number of dominant alleles rather than specific combinations. We use the Binomial Expansion Formula:
Probability = ⁿCᵣ × (p)ʳ × (q)ⁿ^-ʳ
Step 1: Calculate the combinatorial factor (ⁿCᵣ).
Step 2: Plug into the formula.
Frequency of Dominant Alleles in F2 Generation
| Number of Dominant Alleles | Proportion (/64) |
|---|---|
| 0 | 1 |
| 1 | 6 |
| 2 | 15 |
| 3 | 20 |
| 4 | 15 |
| 5 | 6 |
| 6 | 1 |
[!CAUTION] Common Pitfall / Trap: Students often get paralyzed trying to list all 64 possible genotypes (e.g., AABBCc, AaBBCc, AABbCc, etc.). In polygenic traits, AABbcc produces the exact same phenotype as AaBbCc (both have 3 dominant alleles). Applying binomial expansion saves an immense amount of time.