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Genetics: Chapter 1 โ€” Mendelian inheritance

Module 01 โ€ข Chapter Study Notes Part of ๐Ÿงฌ Genetics

Chapter 1: Mendelian inheritance

๐Ÿ“Œ Core Learning Outcomes

Explain why blending inheritance cannot store variation. State segregation and independent assortment from the pea crosses. Calculate a simple probability and a one-degree-of-freedom chi-square, and say what the p value is.

1.1 Before Mendel #

The study of Before Mendel provides fundamental principles governing cellular architecture, molecular interactions, and physiological regulation.

๐Ÿ”ฌ Key Illustrated Scientific Figure
Punnett Square Mendel Flowers
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Punnett Square Mendel Flowers

Interpretation & Mechanism: Genetics diagram: Punnett square describing one of Mendel’s crosses, between parents that are heterozygous for the purple/white color alleles.

1.1.1 Blending as a theory of heredity #

Before Mendel, hereditary contributions were often imagined to mix permanently, like paints. An intermediate would not separate again, and each round of mating would erase about half the standing variation. Selection would run out of differences. Evolutionary Biology needs heritable variation; blending removes it. Darwin felt the difficulty and did not have the particulate answer.

The picture also fails a direct observation. A uniform first generation can produce grandchildren as extreme as the original parents. Hidden variation returns, so what is inherited can be concealed for a generation without being diluted.

1.1.2 What a particle changes #

A particle can be hidden without being changed. If one factor is dominant, the hybrid resembles one parent, yet half its gametes still carry the other factor. If neither is fully dominant, the hybrid looks intermediate and still carries both factors intact. Phenotypes may look blended. The factors do not. Parental types reappear because the factors separate when the hybrid makes gametes.

Mendel spoke of factors, dominance, and recessiveness. Johannsen later named the gene, the genotype, and the phenotype; Bateson named genetics. An allele is one alternative form of a gene. Allele, genotype, and phenotype are enough to state every ratio in this block.

1.2 The pea experiments #

1.2.1 Why Pisum worked #

Mendel used the garden pea, Pisum sativum. It self-pollinates, so a line left to itself becomes true-breeding, and the flower can still be emasculated and crossed by hand. Many seeds set in a short generation, and some characters are scored on the seed itself. He followed seven alternatives: seed shape, seed colour, flower colour, pod shape, pod colour, flower position, and stem length. The combinations he published assorted independently. Linkage is Chapter 4.

He published in 1866. The work was largely ignored until 1900, when de Vries, Correns, and von Tschermak reached the same pattern. Correns returns in Chapter 8. Rediscovery was easier once chromosomes and meiosis were known. Cell Biology owns the stages. The parallel used here is only that homologues separate, and so do alleles.

1.2.2 Segregation and independent assortment #

True-breeding tall crossed with true-breeding dwarf gives a tall F1 if tall is dominant. Selfing the F1 gives about three tall to one dwarf. Selfed F2 dwarfs breed true. Of the F2 talls, about one third breed true and two thirds segregate again, so homozygotes and heterozygotes among the dominant class are one to two.

The law of segregation: a diploid carries two alleles; they separate so each gamete carries one; fertilisation joins gametes at random. With T dominant to t, F2 genotypes are 1 TT : 2 Tt : 1 tt, and dominance reads this as 3:1. The monohybrid ratio is a 1:2:1 seen through dominance.

Two characters give 9:3:3:1 when round and yellow are dominant. Each character alone is still 3:1, and combinations appear that neither pure parent had. Independent assortment says alleles of different genes go to gametes independently when the genes are unlinked, or so far apart that crossing over frees them. A dihybrid makes four gametes in equal frequency. The law does not say every pair of genes is free of every other.

1.3 Predicting a cross #

1.3.1 The test cross #

A dominant phenotype hides TT versus Tt. Mate the unknown to tt. All tall offspring are expected if the unknown is TT; half are dwarf if it is Tt. One dwarf offspring is decisive. A short run of tall offspring is not: from a heterozygote the chance of n dominant offspring is (1/2)^n. Ten tall offspring are unlikely, about 1 in 1,024, and still not impossible.

For two genes, Aa Bb ร— aa bb gives 1:1:1:1 if assortment is independent. Every gamete from the dihybrid is visible, which is why mapping crosses are test crosses. Humans cannot be test-crossed. Chapter 11 replaces that design.

1.3.2 The product rule and the sum rule #

Independent events multiply. From Aa Bb with independent assortment, the probability of an ab gamete is (1/2) ร— (1/2) = 1/4. The probability that the first two children of two carriers are both affected, in that order, is (1/4) ร— (1/4) = 1/16.

Exclusive events add. From Aa ร— Aa, the probability a child is homozygous, AA or aa, is 1/4 + 1/4 = 1/2. Adding events that can overlap double-counts the overlap. The probability of one affected and one unaffected child, in either order, is (1/4)(3/4) + (3/4)(1/4) = 3/8. These are statements about chance, not about what a family ought to do.

1.3.3 Chi-square and what a p value means #

Counts are not exact ratios. Chi-square goodness of fit is ฯ‡ยฒ = ฮฃ (O โˆ’ E)ยฒ / E. Dividing by the expected count stops a difference of ten from being treated as equally surprising in a small class and a large one.

๐Ÿ”ฌ Key Illustrated Scientific Figure
Punnett Square Blood Types
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Punnett Square Blood Types

Interpretation & Mechanism: Punnett square of the possible genotypes and phenotypes of children given the genotypes and phenotypes of their mothers (rows) and fathers (columns) shaded by phenotypes (A: amber, B: blue, AB: green and O: grey) by CMG Lee.

Constructed monohybrid, not a published cross. Two heterozygotes give 300 progeny, 210 dominant and 90 recessive. Under 3:1 the expectations are 225 and 75.

ClassOE(O โˆ’ E)ยฒ / E
Dominant2102251
Recessive90753
Total300300ฯ‡ยฒ = 4

Two classes that must sum to 300 give one degree of freedom; the ratio was fixed in advance. The 5% critical value at one degree of freedom is 3.84. Here ฯ‡ยฒ = 4, so p is just under 0.05 (about 0.046), and 3:1 is rejected at the 5% level.

The p value is the probability of a statistic at least this large if the hypothesis is true and only chance is acting. It is not the probability that the hypothesis is true. Rejection does not name the cause: viability, mis-scoring, linkage to a lethal, a different true ratio, or an unlucky sample can all inflate ฯ‡ยฒ. A value below 3.84 would not prove 3:1; it would only fail to reject it. The same test is used later for 9:3:3:1, 2:1, and 1:1:1:1.

๐Ÿ”ฌ Key Illustrated Scientific Figure
Hardyโ€“Weinberg Law - Punnett Square
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Hardyโ€“Weinberg Law – Punnett Square

Interpretation & Mechanism: Hardyโ€“Weinberg law in Punnett square.

1.4 Key terms, summary, and self-check #

Key terms. Blending inheritance; allele; genotype; phenotype; segregation; independent assortment; test cross; product rule; sum rule; chi-square; p value.

Summary. Blending would destroy variation; Mendelian factors stay intact in a hybrid and separate at gamete formation. Segregation gives 1:2:1 genotypically and, with dominance, 3:1 phenotypically. Independent assortment gives 9:3:3:1. A test cross exposes a hidden genotype. Independent events multiply and exclusive events add. Chi-square compares counts with a ratio, and its p value is the chance of a departure at least that large if the hypothesis is true.

Self-check.

1. Why does one recessive test-cross offspring prove heterozygosity, while a small all-dominant progeny does not prove homozygosity?

2. For 290 dominant and 110 recessive F2 plants tested against 3:1, what are O and E, and how many degrees of freedom are there?

3. Use the product rule to give the probability of an ab gamete from Aa Bb under independent assortment.

4. Among dominant F2 plants from a monohybrid self, what fraction are heterozygous?

5. What does a p value of 0.20 for a 3:1 hypothesis mean, and what does it not mean?

๐Ÿง  High-Yield Memory Hacks & Visual Mnemonics

Unforgettable Memory Hacks & Diagrams for Chapter 1

Retain complex pathways, biological sequences, and tricky examiner traps effortlessly with these visual mental pegs, funny stories, and high-yield mnemonics.

Epistasis Ratios Derived from 9:3:3:1
โ€œ
9:3:4 Recessive, 12:3:1 Dominant, 9:7 Complementary, 15:1 Duplicate
๐ŸŽญ
Picture This (Mental Hook): Imagine a modified casino board where 9:3:3:1 gets rearranged by gene interactions! When Recessive epistasis strikes, the last two slots merge (9:3:4 Labrador coat color). When Dominant epistasis strikes, the first two slots merge (12:3:1 Summer squash)!
๐Ÿ“Š Visual Diagram & Mental Flow:
โ”Œโ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”ฌโ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”ฌโ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”
โ”‚ Interaction Type        โ”‚ F2 Ratio     โ”‚ Classical Example                                  โ”‚
โ”œโ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”ผโ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”ผโ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”ค
โ”‚ Mendelian Dihybrid      โ”‚ 9:3:3:1      โ”‚ Independent assortment (Pea seed shape & color)    โ”‚
โ”‚ Recessive Epistasis     โ”‚ 9:3:4        โ”‚ Labrador retriever coat (ee masks B black / b brownโ”‚
โ”‚ Dominant Epistasis      โ”‚ 12:3:1       โ”‚ Fruit color in Summer squash (Cucurbita pepo)      โ”‚
โ”‚ Duplicate Recessive     โ”‚ 9:7          โ”‚ Flower color in Sweet Pea (Lathyrus odoratus)      โ”‚
โ”‚ (Complementary Genes)   โ”‚              โ”‚ (Both enzymes needed for purple anthocyanin)       โ”‚
โ”‚ Duplicate Dominant      โ”‚ 15:1         โ”‚ Seed capsule shape in Shepherd's purse (Capsella)  โ”‚
โ”‚ Dominant & Recessive    โ”‚ 13:3         โ”‚ Feather color in fowl (Inhibitory gene interaction)โ”‚
โ””โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”ดโ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”ดโ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”€โ”˜
๐Ÿ” The Decoded Sequence:
9:3:4
Recessive Epistasis Homozygous recessive at one locus masks expression of both alleles at second locus
12:3:1
Dominant Epistasis Dominant allele at one locus produces phenotype regardless of second locus
9:7
Complementary Genes At least one dominant allele at BOTH loci required for wildtype phenotype
โš ๏ธ
Examiner Trap / Never Forget: Notice that in ALL epistatic modifications, the sum of the ratio numbers ALWAYS equals 16 (e.g. 9+3+4=16; 12+3+1=16; 9+7=16; 15+1=16)!
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