DNA, genes, mitosis & meiosis

What you will learn

describe the structure of DNA, chromosomes, genes and alleles,
distinguish mitosis (body cell division) from meiosis (gamete formation),
define dominant, recessive, genotype, phenotype and homozygous/heterozygous,
predict offspring ratios from a monohybrid Punnett square,
explain the inheritance of sex-linked traits.

Worked example 0 Real-world example: predicting eye colour in a family

Brown eye allele $B$ is dominant over blue allele $b$ . Both parents are heterozygous ( $Bb$ ). What fraction of their children are expected to have blue eyes?

Each parent can pass on either $B$ or $b$ with equal probability.
Draw a 2 by 2 Punnett square: possible combinations are $BB$ , $Bb$ , $Bb$ , $bb$ .
Genotype ratio: $1 : 2 : 1$ . Phenotype ratio (brown : blue) $= 3 : 1$ .
Expected fraction with blue eyes: $\dfrac{1}{4}$ , or $25\%$ .

Key idea: a cross between two heterozygotes gives the classic $3 : 1$ phenotypic ratio whenever one allele is fully dominant.

1. DNA, chromosomes, genes and alleles

DNA (deoxyribonucleic acid) is a double helix held together by base pairs. Adenine pairs with thymine (A-T) and guanine pairs with cytosine (G-C). The sequence of bases is the genetic code.

Gene: a length of DNA that codes for a particular protein or trait.
Chromosome: a long DNA molecule wrapped around proteins. Humans have $23$ pairs (46 in total).
Allele: one of the possible versions of a gene (e.g. $B$ for brown eyes, $b$ for blue).
Genotype: the pair of alleles an individual carries ( $BB$ , $Bb$ or $bb$ ).
Phenotype: the observable trait produced by the genotype.

DNA double helix with complementary base pairing (A-T, G-C).

2. Mitosis vs meiosis

Mitosis produces two genetically identical daughter cells. It is used for growth, repair and asexual reproduction. Each daughter has the full $46$ chromosomes.

Meiosis produces four genetically different gametes (sperm or egg cells), each with $23$ chromosomes — half the normal number. When a sperm fertilises an egg the full set is restored in the offspring.

Feature	Mitosis	Meiosis
Number of divisions	1	2
Daughter cells	2	4
Chromosomes per cell	46 (diploid)	23 (haploid)
Genetic variation	identical copies	new combinations
Purpose	growth, repair	sexual reproduction

Worked example 1 Counting chromosomes

A human skin cell has $46$ chromosomes. After mitosis, how many chromosomes does each daughter cell have? After meiosis?

Mitosis: DNA is copied, then divided equally. Each daughter cell has $46$ chromosomes.
Meiosis: chromosomes are copied once but the cell divides twice, so each gamete has $\dfrac{46}{2} = 23$ chromosomes.

At fertilisation, a sperm ( $23$ ) joins an egg ( $23$ ) to give a zygote with $46$ chromosomes — the full diploid set.

3. Mendelian inheritance and Punnett squares

A Punnett square is a grid that shows all possible allele combinations in the offspring. The rows give one parent’s gametes, the columns give the other’s.

Worked example 2 Monohybrid cross, one heterozygous parent

In pea plants, tall ( $T$ ) is dominant over short ( $t$ ). Cross a heterozygous tall plant ( $Tt$ ) with a short plant ( $tt$ ). What are the genotype and phenotype ratios of the offspring?

$Tt$ parent produces gametes $T$ and $t$ (each $\tfrac{1}{2}$ ).
$tt$ parent produces only $t$ gametes.
Punnett square gives offspring: $Tt, Tt, tt, tt$ .
Genotype ratio $1\,Tt : 1\,tt$ . Phenotype ratio (tall : short) $= 1 : 1$ .

So about half the offspring are tall, half are short.

Worked example 3 Carriers of a recessive disease

Cystic fibrosis is caused by a recessive allele $f$ . Two heterozygous carriers ( $Ff$ ) have a child. Find the probability the child (a) has the disease and (b) is an unaffected carrier.

Punnett square: $FF, Ff, Ff, ff$ .
(a) Has disease $=$ genotype $ff$ : probability $\dfrac{1}{4}$ , or $25\%$ .
(b) Carrier $=$ genotype $Ff$ : probability $\dfrac{2}{4} = \dfrac{1}{2}$ , or $50\%$ .

Key idea: two unaffected carriers can produce an affected child because each has a one-in-two chance of passing on the faulty allele.

4. Sex-linked inheritance

Sex chromosomes are $X$ and $Y$ . Females are $XX$ , males are $XY$ . Genes on the $X$ chromosome show sex-linked inheritance: because males have only one $X$ , a single recessive allele is enough to express the trait.

Worked example 4 Colour blindness (X-linked recessive)

The colour-blind allele $X^c$ is recessive. A carrier mother ( $X^C X^c$ ) has a child with a normal-vision father ( $X^C Y$ ). Find the probability that a son is colour-blind.

Mother’s gametes: $X^C$ or $X^c$ (each $\tfrac{1}{2}$ ).
Father’s gametes: $X^C$ or $Y$ (each $\tfrac{1}{2}$ ).
Sons receive $Y$ from father. So son genotypes are $X^C Y$ or $X^c Y$ , each with probability $\tfrac{1}{2}$ .
Probability a son is colour-blind $= \dfrac{1}{2}$ . Daughters all receive $X^C$ from the father and are unaffected, though half are carriers.

Key idea: X-linked recessive conditions are far more common in males because they only need one copy of the allele to express it.

Practice: Year 10

Fluency

Vocabulary and structure

Name the four bases of DNA and state which pairs with which.
How many chromosomes are in a normal human body cell? A normal gamete?
Define: gene, allele, genotype, phenotype.
A pea plant has genotype $Tt$ . Is it homozygous or heterozygous?
List two purposes of mitosis in the human body.
State one key difference between mitosis and meiosis.

Fluency

Punnett squares

Cross $Bb \times Bb$ (brown $B$ dominant, blue $b$ ). State the phenotype ratio.
Cross $BB \times bb$ . State the genotype of all offspring and their phenotype.
Cross $Tt \times tt$ (tall dominant). State the expected phenotype ratio.
In guinea pigs, black fur ( $B$ ) is dominant over white ( $b$ ). What is the probability that two heterozygous parents have a white offspring?
Two parents both have attached earlobes (recessive, genotype $ee$ ). What earlobe phenotype will all their children have? Explain.

Reasoning

Applied inheritance

In humans, widow’s peak hairline ( $W$ ) is dominant over straight ( $w$ ). A woman with a widow’s peak whose mother had a straight hairline marries a man with a straight hairline. What is her genotype? What fraction of their children are expected to have a widow’s peak?
Two carriers of cystic fibrosis ( $Ff$ ) plan to have four children. What is the expected number of affected children? Why might the actual number differ?
Explain why a cross between two heterozygotes produces a $3 : 1$ phenotypic ratio but a $1 : 2 : 1$ genotypic ratio.
A colour-blind father and a non-carrier mother have children. Can any of their daughters be colour-blind? Can any of their sons? Justify with a Punnett square.
A farmer crosses two red-flowered plants and gets $75$ red and $24$ white offspring. Deduce the genotypes of the parents and write the Punnett square.

Problem solving

Meiosis and variation

Explain how meiosis, followed by random fertilisation, produces genetic variation in offspring even though the parents’ genes do not change.
A cell with $8$ chromosomes undergoes meiosis. How many chromosomes are in each gamete? How many gametes are produced from one starting cell?
Haemophilia is X-linked recessive. A carrier mother and an unaffected father have a son. What is the probability that the son has haemophilia? What is the probability a daughter is a carrier?
In snapdragons, red ( $R$ ) and white ( $r$ ) alleles show incomplete dominance: $Rr$ gives pink. Cross $Rr \times Rr$ and give the phenotype ratio.

Challenge

Reasoning

Harder reasoning

A couple has three daughters already. They ask: “What is the probability our next child is a girl?” Explain why the answer is still $\tfrac{1}{2}$ , not $\tfrac{1}{16}$ . Link your answer to the independence of meiosis events.
Huntington’s disease is autosomal dominant. A man whose father had Huntington’s has not yet developed symptoms. If his mother is unaffected ( $hh$ ), what is the probability that he carries the allele? If he does, what is the probability each of his children inherits it?
A dihybrid cross involves two genes at once. If pea colour (yellow $Y$ dominant, green $y$ ) and shape (round $R$ dominant, wrinkled $r$ ) assort independently, predict the phenotype ratio of offspring from a cross $YyRr \times YyRr$ . (Hint: $9 : 3 : 3 : 1$ .) Justify the $9$ category.
Evolutionary biologists argue that sexual reproduction, though costly, persists because meiosis generates variation. Explain how (i) independent assortment of chromosomes and (ii) random fertilisation combine to produce a vast number of possible offspring genotypes from a single couple.

Answers

Answer key

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Year 10 answers

Fluency

Vocabulary and structure

Adenine, thymine, guanine, cytosine. A pairs with T; G pairs with C.
Body cell: $46$ chromosomes. Gamete: $23$ chromosomes.
Gene — a length of DNA coding for a trait. Allele — one version of a gene. Genotype — the pair of alleles carried. Phenotype — the observable trait.
Heterozygous (two different alleles).
Growth and repair of tissues; replacing old cells (e.g. skin, blood).
Mitosis produces two identical diploid cells; meiosis produces four genetically different haploid gametes.

Fluency

Punnett squares

$3 : 1$ brown to blue.
All offspring $Bb$ ; all brown.
$1 : 1$ tall to short.
$\dfrac{1}{4}$ , or $25\%$ .
All children will have attached earlobes. Both parents are $ee$ , so every child must inherit $e$ from each parent.

Reasoning

Applied inheritance

Her mother was $ww$ , so the woman inherited $w$ from mother and $W$ from her widow’s-peak parent; genotype $Ww$ . Cross $Ww \times ww$ : $\dfrac{1}{2}$ widow’s peak, $\dfrac{1}{2}$ straight.
Expected $\dfrac{1}{4} \times 4 = 1$ affected child. The actual number can differ because each child is an independent event; outcomes follow a binomial distribution, not a guaranteed quota.
Offspring genotypes $BB : Bb : bb = 1 : 2 : 1$ . Because $BB$ and $Bb$ share the dominant phenotype, the three “brown” outcomes combine, giving $3 : 1$ .
Father is $X^c Y$ , mother is $X^C X^C$ . All daughters are $X^C X^c$ (carriers, not colour-blind). All sons are $X^C Y$ (not colour-blind).
Ratio $\approx 3 : 1$ suggests both parents are heterozygous $Rr$ . Punnett: $RR, Rr, Rr, rr$ , giving $3$ red $: 1$ white.

Problem solving

Meiosis and variation

Meiosis shuffles maternal and paternal chromosomes by independent assortment and crossing over, producing gametes with new combinations. Random fertilisation then pairs two such gametes, so each child represents one of millions of possible genotypes.
$4$ chromosomes per gamete; $4$ gametes from one starting cell.
Son has haemophilia with probability $\dfrac{1}{2}$ . Daughter is a carrier with probability $\dfrac{1}{2}$ (the other $\dfrac{1}{2}$ are $X^H X^H$ ).
$1 : 2 : 1$ red : pink : white.

Reasoning

Challenge

Each conception is independent; meiosis in the father randomly produces an $X$ -bearing or $Y$ -bearing sperm with probability $\tfrac{1}{2}$ . Previous children do not change the odds for the next.
His father must have been $Hh$ (or he could not pass the allele), so the son’s probability of being $Hh$ is $\dfrac{1}{2}$ . If he is $Hh$ , each child has probability $\dfrac{1}{2}$ of inheriting $H$ .
Ratio $9 : 3 : 3 : 1$ (both dominant : yellow round; yellow wrinkled; green round; green wrinkled). The $9$ comes from $\dfrac{3}{4} \times \dfrac{3}{4} = \dfrac{9}{16}$ having at least one dominant allele for each gene.
(i) With $23$ chromosome pairs, independent assortment gives $2^{23} \approx 8.4 \times 10^6$ gamete combinations per parent. (ii) Random fertilisation squares this: $\approx 7 \times 10^{13}$ possible zygote genotypes, before crossing over. Variation fuels adaptation to changing environments.

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