Genetics for NEET: the ratios, the disorders and the questions that repeat

Mendel's laws and the standard ratios

Dominance, segregation and independent assortment, demonstrated on the garden pea and its seven contrasting characters. Learn the monohybrid and dihybrid ratios with their genotypic versions, and the test cross that tells a homozygote from a heterozygote.

When the ratio changes

Incomplete dominance gives 1:2:1 for both genotype and phenotype; codominance shows both alleles, as in the AB blood group; multiple alleles, pleiotropy and polygenic inheritance each bend the expected numbers. Questions usually describe the situation and expect you to name it.

Linkage and recombination

Morgan's work on Drosophila showed that genes on the same chromosome do not assort independently. Recombination frequency measures how often they separate, and that frequency became the basis for mapping genes along a chromosome.

Sex determination and pedigrees

XX–XY in humans, XX–XO in grasshoppers, ZZ–ZW in birds and haplodiploidy in honeybees. Pedigree questions almost always reduce to one decision: is the trait autosomal or X-linked, dominant or recessive.

Mendelian and chromosomal disorders

Haemophilia and colour blindness are X-linked recessive; sickle-cell anaemia, thalassaemia and phenylketonuria are autosomal recessive. Chromosomal disorders come from aneuploidy: trisomy 21 in Down syndrome, XXY in Klinefelter syndrome, XO in Turner syndrome.

DNA as the genetic material

Griffith's transforming principle, the Avery–MacLeod–McCarty follow-up and the Hershey–Chase experiment with radioactively labelled bacteriophages. Know what each experiment could and could not conclude — that distinction is the question.

Replication, transcription and translation

Semiconservative replication, proved by Meselson and Stahl in E. coli; the transcription unit and its promoter and terminator; the genetic code and its properties; and the lac operon as the standard example of regulation.

Monohybrid cross

3 : 1 phenotypic · 1 : 2 : 1 genotypic

From a heterozygote selfed. The genotypic ratio is the one questions hide behind a phenotype count.

Dihybrid cross

9 : 3 : 3 : 1

Only when the two genes assort independently and both show complete dominance.

Test cross

1 : 1 for one gene · 1 : 1 : 1 : 1 for two

Crossing with the homozygous recessive reveals the unknown genotype directly.

Incomplete dominance

1 : 2 : 1 phenotypic = genotypic

The heterozygote has its own intermediate phenotype, so phenotype and genotype ratios coincide.

Gamete types

2ⁿ for n heterozygous gene pairs

AaBb gives 4 gamete types; AaBbCc gives 8. A quick check before drawing any square.

Chargaff's rules

A = T · G = C · purines = pyrimidines

So knowing the share of one base fixes the share of its partner.

B-DNA geometry

10 base pairs per turn · pitch 3.4 nm · 0.34 nm per base pair

The three numbers are related: 10 × 0.34 nm = 3.4 nm.

The genetic code

64 codons · 61 code for amino acids · UAA, UAG, UGA stop

AUG is both the initiator codon and the codon for methionine. The code is degenerate but unambiguous.

Sex determination

Human XX/XY · grasshopper XX/XO · birds ZZ/ZW

In birds the female is the heterogametic sex — the reversal is the whole question.

Chromosomal disorders

Trisomy 21 → 47 · XXY → 47 · XO → 45

Down, Klinefelter and Turner syndromes respectively, all from non-disjunction.

Sickle-cell anaemia

Glu → Val at the sixth position of the β-globin chain

A single base substitution, inherited as an autosomal recessive trait.

The lac operon

i gene repressor · z, y, a structural genes

Lactose acts as the inducer; z gives β-galactosidase, y permease and a transacetylase.

Probability · dihybrid cross

Two plants of genotype AaBb are crossed, with both genes unlinked and completely dominant. What fraction of the offspring is homozygous recessive for both genes?

1. Treat the genes separately. From Aa × Aa, the chance of aa is 1/4.
2. From Bb × Bb, the chance of bb is also 1/4.
3. Because the genes assort independently, multiply: 1/4 × 1/4 = 1/16.

Answer: One in sixteen — the single "1" at the end of the 9:3:3:1 ratio.

Reasoning · multiple alleles

A mother has blood group O and the father has blood group AB. Which blood groups can their children have, and which are impossible?

1. Write the genotypes: the mother is ii, and the father is IᴬIᴮ.
2. List the gametes: the mother gives only i; the father gives Iᴬ or Iᴮ.
3. Combine them: children are Iᴬi or Iᴮi.

Answer: The children can be group A or group B only. Neither O nor AB is possible, because every child receives one i from the mother and one dominant allele from the father.

Assertion and reason · sex linkage

Assertion: a colour-blind daughter must have a colour-blind father. Reason: colour blindness is an X-linked recessive condition.

1. A daughter inherits one X chromosome from each parent.
2. For a recessive X-linked trait to show in a daughter, both her X chromosomes must carry the allele.
3. Her father contributes his only X chromosome, so if it carries the allele he has no second X to mask it and is colour-blind himself.

Answer: Both statements are true, and the reason explains the assertion. The mother, by contrast, may be an unaffected carrier.