Principles of Inheritance and Variation

Haemoglobin has two $\alpha$ and two $\beta$ polypeptide chains.
In sickle cell anaemia, a point mutation in the $\beta$-globin gene (GAG $\rightarrow$ GUG in the codon) replaces glutamic acid by valine at the sixth position of the $\beta$ chain.
The resulting HbS polymerises at low oxygen, making the RBCs sickle shaped.
So sickle cell anaemia is due to change of an amino acid in the $\beta$-chain of haemoglobin.

Mendel's law of independent assortment holds for genes located on different (non-homologous) chromosomes, which separate independently during meiosis.
Genes located close together on the same chromosome are linked; they tend to be inherited together and do not assort independently.
So independent assortment does not take place when genes are linked and located on the same chromosome.

Maleness in humans is determined by the SRY gene (sex-determining region of Y), located near the tip of the Y chromosome's short arm.
During meiosis in the father, an abnormal crossing over between the X and Y chromosomes can transfer the SRY region from Y to X.
An XX zygote that receives an X carrying SRY develops as a male (XX male), while an XY zygote whose Y has lost SRY develops as a female (XY female).
So these conditions occur due to transfer of segments between the X and Y chromosomes.

In killer strains of Paramecium aurelia, the killer trait is due to kappa particles, which are present in the cytoplasm (they are symbiotic bacteria).
Kappa particles produce a toxin (paramecin) that kills sensitive strains. They are inherited through the cytoplasm, not through nuclear genes, though a nuclear gene K is needed to maintain them (studied by Sonneborn).
So extranuclear (cytoplasmic) inheritance occurs in killer Paramecium.

The R gene of pea codes for starch branching enzyme (SBE1), which converts amylose into amylopectin.
In rr plants, a mutation (a transposon insertion) makes this enzyme non-functional. Less starch is made and more sugar accumulates in the seed.
The extra sugar draws in water by osmosis while the seed develops; when the seed matures and dries, it loses this water and shrinks, so it becomes wrinkled.
So wrinkled seeds were due to the absence of the starch branching enzyme.

Complementary genes are two non-allelic dominant genes that produce a character only when both are present together (e.g. purple flower colour in sweet pea, studied by Bateson and Punnett).
In the $F_2$ of a dihybrid cross: 9 $A\_B\_$ show the character, while 3 $A\_bb$ + 3 $aaB\_$ + 1 aabb (= 7) do not.
So the ratio is 9 : 7.
(9 : 3 : 4 is recessive epistasis and 12 : 3 : 1 is dominant epistasis.)

Complementary genes are two non-allelic dominant genes that interact to produce a new phenotype only when both are present together; neither alone produces it.
Example: in sweet pea, purple flowers appear only when both C and P are present; CCpp and ccPP are both white.
Epistasis is when one gene masks the effect of another gene, and polygenes add up their effects (quantitative inheritance).
So such genes are called complementary genes.

In co-dominance, both alleles of a gene express themselves fully in the heterozygote; neither masks the other.
Example: in the ABO blood group, $I^A$ and $I^B$ are both expressed in $I^AI^B$ (blood group AB); in cattle, red and white give roan.
In dominance only the dominant allele is expressed, and in incomplete dominance the heterozygote is intermediate.
So when both alleles express themselves together, it is called co-dominance.

With complete linkage, the heterozygote AB/ab forms only parental gametes: AB and ab.
The homozygous recessive parent ab/ab forms only ab gametes.
Progeny: AB $\times$ ab = AB/ab (AaBb) and ab $\times$ ab = ab/ab (aabb), in a 1 : 1 ratio.
(If crossing over occurred, a few recombinants Aabb and aaBb would also appear.)
So the progeny genotypes are AaBb and aabb.

In humans, the probability of each child being a son is $\dfrac{1}{2}$, because the father's sperm carries X or Y in equal numbers.
Each birth is independent, so the probability of four sons in four births is $\left(\dfrac{1}{2}\right)^4$.
$= \dfrac{1}{16}$

In mammals, only one X chromosome remains active in each cell; all extra X chromosomes are condensed and inactivated as Barr bodies (Lyon hypothesis).
Number of Barr bodies = number of X chromosomes $- 1$.
For XXXX: $4 - 1 = 3$ Barr bodies.
(A normal XX female has 1 Barr body, and a normal male has none.)