Chapter 4: Principles of Inheritance and Variation (Genetics and Evolution)

• Pea plants have many clear contrasting characters (tall/dwarf, round/wrinkled seeds, yellow/green seeds, etc.).
• Flowers are bisexual and naturally self-pollinating, so pure lines are easy to get; they can also be cross-pollinated artificially.
• It is an annual plant with a short life cycle, so many generations can be studied quickly.
• It produces a large number of seeds, giving statistically reliable results.
• It is easy to grow.

Many contrasting characters, self-pollination (pure lines) with easy artificial crossing, short life cycle and many seeds.

(a) A dominant allele expresses itself in both homozygous (TT) and heterozygous (Tt) conditions. A recessive allele expresses itself only in the homozygous condition (tt); in the heterozygote it is masked.

(b) Homozygous: both alleles of a gene are identical (TT or tt); breeds true. Heterozygous: the two alleles are different (Tt); does not breed true.

(c) Monohybrid cross: parents differ in one pair of contrasting characters (e.g. tall × dwarf); F2 phenotypic ratio 3 : 1. Dihybrid cross: parents differ in two pairs of characters (e.g. round yellow × wrinkled green); F2 ratio 9 : 3 : 3 : 1.

(a) Expressed in heterozygote vs only in homozygote. (b) Same alleles vs different alleles. (c) One character pair (3:1) vs two pairs (9:3:3:1).

Number of gamete types $=2^n$, where $n$ is the number of heterozygous loci (assuming independent assortment): $2^4=16$.

16 types of gametes.

Law of dominance: characters are controlled by discrete units called factors, which occur in pairs; when the two factors of a pair are dissimilar, one (dominant) expresses itself and the other (recessive) is masked.

Cross: pure tall (TT) × pure dwarf (tt).

• F1: all Tt, all tall. The dwarf factor is present but not expressed, so T is dominant over t.
• F1 selfed (Tt × Tt) gives F2: TT : Tt : tt = 1 : 2 : 1, i.e. 3 tall : 1 dwarf. The dwarf character reappears in F2, proving that the recessive factor was only hidden, not lost.

TT × tt gives all tall Tt in F1 (dwarf masked); F2 shows 3 tall : 1 dwarf.

A test cross is a cross between an organism showing the dominant phenotype (genotype unknown) and the homozygous recessive parent, to find out the unknown genotype.

Design: a tall pea plant (TT or Tt?) is crossed with a dwarf plant (tt).

• If all offspring are tall, the plant was TT (homozygous).
• If tall and dwarf appear in a 1 : 1 ratio, the plant was Tt (heterozygous).

Crossing a dominant-phenotype individual with the homozygous recessive: all dominant offspring means homozygous; 1:1 means heterozygous.

Take a tall heterozygous male (Tt).

Case 1: homozygous dominant female (TT) × Tt

All offspring are tall (genotypes TT : Tt = 1 : 1).

Case 2: homozygous recessive female (tt) × Tt

Tall : dwarf = 1 : 1.

TT female × Tt: all show the dominant trait. tt female × Tt: dominant : recessive = 1 : 1.

The two genes assort independently.

• Height: Tt × Tt gives tall $\frac{3}{4}$, dwarf $\frac{1}{4}$.
• Seed colour: Yy × yy gives yellow $\frac{1}{2}$, green $\frac{1}{2}$.

(a) Tall and green $=\frac{3}{4}\times \frac{1}{2}=\frac{3}{8}$

(b) Dwarf and green $=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$

(a) 3/8. (b) 1/8.

If the two genes are completely linked (on the same chromosome, no crossing over), they are inherited together like a single gene. A cross of AB/ab × AB/ab gives only the parental combinations, in a ratio of about 3 : 1 (3 with both dominant traits : 1 with both recessive traits), instead of the 9 : 3 : 3 : 1 of independent assortment.

If linkage is incomplete, parental types are still in large majority, with a small number of recombinant types formed by crossing over.

Completely linked genes behave as one: about 3 : 1 parental types only (incomplete linkage adds a few recombinants).

• Worked with the fruit fly Drosophila melanogaster and gave experimental proof of the chromosomal theory of inheritance.
• Discovered linkage (genes on the same chromosome tend to be inherited together) and recombination (new combinations by crossing over).
• Showed that the strength of linkage depends on the distance between genes; his student Sturtevant used recombination frequency to map genes on chromosomes.
• Studied sex-linked inheritance (e.g. white eye colour in Drosophila).

Proved the chromosomal theory using Drosophila, discovered linkage and recombination, and led to gene mapping.

Pedigree analysis is the study of the inheritance of a trait over several generations of a family, using a chart (pedigree) with standard symbols.

Uses:

• Finds whether a trait or disorder is dominant or recessive, and autosomal or sex-linked.
• Traces the origin and passage of genetic disorders (e.g. haemophilia, sickle-cell anaemia) in a family.
• Helps genetic counselling: predicting the chance of a disorder in future children.

Studying a trait's inheritance through generations of a family; it shows the pattern of inheritance and helps predict disorders (genetic counselling).

Humans have the XX–XY type of sex determination. Of the 23 pairs of chromosomes, 22 are autosomes and one pair is the sex chromosomes: XX in females and XY in males.

• All ova carry an X chromosome.
• Half the sperms carry X and half carry Y.
• Egg + X sperm → XX (female); egg + Y sperm → XY (male).

So the sex of the child is decided by the sperm, and the chance of a boy or girl is 50 : 50.

XX–XY system: the sperm decides; X sperm gives a girl (XX), Y sperm a boy (XY).

A child with blood group O is $ii$, so each parent must carry the $i$ allele.

• Father (A): $I^{A}i$
• Mother (B): $I^{B}i$

The other children may be $I^A{I}^B$ (AB), $I^{A}i$ (A), $I^{B}i$ (B) or $ii$ (O), each with a chance of 1 in 4.

Father $I^{A}i$, mother $I^{B}i$; offspring can be $I^A{I}^B$, $I^{A}i$, $I^{B}i$ or $ii$ (AB, A, B, O).

(a) Co-dominance: both alleles of a heterozygote express themselves fully and together. Example: ABO blood groups: $I^A$ and $I^B$ are co-dominant, so $I^A{I}^B$ has both A and B antigens (blood group AB).

(b) Incomplete dominance: neither allele is completely dominant, so the heterozygote shows an intermediate phenotype. Example: in snapdragon (Antirrhinum), red (RR) × white (rr) gives pink (Rr) F1; the F2 ratio is 1 red : 2 pink : 1 white (phenotype and genotype ratios are the same).

(a) Both alleles expressed (blood group AB). (b) Blended intermediate (pink snapdragon from red × white).

A point mutation is a change in a single base pair of DNA. Example: sickle-cell anaemia, caused by the substitution of A by T in the gene for the β-globin chain, so that glutamic acid is replaced by valine at the sixth position (GAG → GUG codon).

A change in a single base pair; e.g. sickle-cell anaemia (Glu → Val at position 6 of β-globin).

Walter Sutton and Theodor Boveri (1902). It was later verified experimentally by T.H. Morgan.

Sutton and Boveri.

• Sickle-cell anaemia (autosomal recessive): red blood cells become sickle-shaped under low oxygen; anaemia, pain, poor blood flow.
• Phenylketonuria (autosomal recessive): lack of the enzyme that converts phenylalanine to tyrosine; phenylalanine builds up, causing mental retardation; excreted in urine.
• Down's syndrome (trisomy of chromosome 21): short stature, small round head, furrowed tongue, partly open mouth, palm crease, retarded physical and mental development.

For example, sickle-cell anaemia (sickled RBCs, anaemia) and phenylketonuria (phenylalanine build-up, mental retardation); also Down's syndrome.