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IMAT Biology · Genetics and Inheritance

Linkage, Gene Maps, Variation and Non-disjunction

Genes on the same chromosome travel together unless crossing over separates them; how often that happens maps the genes, and errors in meiosis change the chromosome number.

Why this matters

Gene order from crossover values was asked in 2011 and 2018. Other older questions asked why a brother and sister differ (2018), what mitosis does to two linked genes (2016), and whether a faulty sperm with the SRY gene could give a female with an extra sex chromosome (2021).

Concept 1 of 4: Linked genes and recombination frequency

Two genes on the same chromosome are physically joined, so they tend to enter the same gamete. They can be separated only when homologous chromosomes swap segments by crossing over between them. The further apart the genes are, the more often a crossover falls between them, so the share of recombinant offspring measures their distance.

Definition

  • Linked genes are on the same chromosome and do not assort independently.
  • Crossing over happens in prophase I of meiosis, between non-sister chromatids of homologous chromosomes, at points called chiasmata.
  • In a test cross of a double heterozygote, parental combinations are the majority; recombinant (new) combinations are the minority.
  • The recombination frequency (crossover value) is the percentage of recombinant offspring. It is at most 50%; 50% means the genes behave as unlinked.
  • Mitosis does not separate linked alleles or homologues: both daughter cells get the same genotype as the parent cell.

Recombination frequency

RF=number of recombinant offspringtotal offspring×100%RF = \frac{\text{number of recombinant offspring}}{\text{total offspring}} \times 100\%
  • RFRFrecombination frequency (crossover value), in %

Worked example

In fruit flies, grey body (GG) and long wings (LL) are dominant to black body (gg) and vestigial wings (ll). A fly heterozygous for both genes, which received GG and LL together from one parent, is test-crossed with a ggllggll fly. Offspring: 420 grey long, 405 black vestigial, 88 grey vestigial, 87 black long. Are the genes linked, and what is the recombination frequency?
Practice this conceptself-check · 4 quick reps

Crossing over happens inside one parent's meiosis

Crossing over is between the two homologous chromosomes of one person, during prophase I as that person makes gametes. It is never between a chromosome from the sperm and one from the egg at fertilisation, and it does not happen in the zygote's mitotic divisions.

Concept 2 of 4: Building a gene map from crossover values

If recombination frequency measures distance, a set of pairwise distances is like the distances between towns on one straight road. The two towns farthest apart must be at the two ends. Every other town is then placed by its distance from one end, and the remaining distances check the answer.

Definition

  • One map unit (one centimorgan, cM) corresponds to a recombination frequency of 1%.
  • Distances along one chromosome add: if C lies between A and B, then AC+CB=ABAC + CB = AB.
  • Method: (1) find the largest value; those two genes are at the ends. (2) Place each other gene by its distance from one end. (3) Check every given value.
  • Over long distances double crossovers go uncounted, so a large measured value can fall slightly short of the sum of the smaller ones.

Distances add along a chromosome

d(A,B)=d(A,C)+d(C,B)when C lies between A and Bd(A,B) = d(A,C) + d(C,B) \quad \text{when C lies between A and B}
  • ddmap distance, in map units (= % recombination)

Worked example

Crossover values for four genes on one chromosome: J and M 26%, J and K 12%, K and L 5%, L and M 9%, K and M 14%. Find the gene order.
Practice this conceptself-check · 3 quick reps

Concept 3 of 4: Meiosis and fertilisation as sources of genetic variation

Brothers and sisters share parents but are not identical. Each parent's meiosis shuffles chromosomes in two ways: whole chromosomes are dealt out at random, and crossing over mixes alleles within chromosomes. Then any sperm can meet any egg. Multiplying these choices gives an enormous number of possible children.

Definition

Three sources of variation in sexual reproduction (beyond new mutations):

  • Independent assortment: homologous pairs line up in random orientation at the equator in metaphase I. With haploid number nn there are 2n2^n chromosome combinations per gamete.
  • Crossing over in prophase I creates new combinations of alleles on each chromosome.
  • Random fertilisation: any gamete of one parent can fuse with any gamete of the other.
  • In humans (n=23n = 23) independent assortment alone gives 223≈8.42^{23} \approx 8.4 million kinds of gamete, and about 7×10137 \times 10^{13} combinations for a zygote.
  • Mutation is the original source of new alleles.

Chromosome combinations from independent assortment

gametes=2nzygotes=2n×2n\text{gametes} = 2^{n} \qquad \text{zygotes} = 2^{n} \times 2^{n}
  • nnhaploid chromosome number (number of homologous pairs)

Worked example

An animal has a diploid number 2n=82n = 8. Ignoring crossing over, how many chromosome combinations can its gametes have, and how many combinations are possible for a zygote from two such parents?
Practice this conceptself-check · 3 quick reps

Variation between siblings is made before fertilisation

The differences between a brother and a sister come from each parent's meiosis (random orientation in metaphase I, crossing over in prophase I) and from which sperm meets which egg. Nothing in the zygote's own mitotic divisions reshuffles alleles.

Concept 4 of 4: Non-disjunction and aneuploidy: Down, Turner and Klinefelter syndromes

In meiosis, each pair of chromosomes must split cleanly between the gametes. If a pair fails to separate, one gamete gets both copies and another gets none. When such a gamete is fertilised, the zygote has one chromosome too many or one too few, in every cell of the body.

Definition

  • Non-disjunction: homologous chromosomes fail to separate in anaphase I, or sister chromatids fail to separate in anaphase II.
  • The result is gametes with n+1n + 1 and n−1n - 1 chromosomes. Fertilisation by a normal gamete gives a trisomy (2n+1=472n + 1 = 47) or a monosomy (2n−1=452n - 1 = 45). This is aneuploidy.
  • Down syndrome is most often due to non-disjunction in the mother's meiosis, and its risk rises with the mother's age.
  • Sex is still decided by SRY: any karyotype with a Y (and working SRY) develops as male.
  • A karyotype (the chromosomes photographed and arranged in pairs) shows these changes; it is used, for example, in prenatal diagnosis.
ConditionKaryotypeSexMain features
Down syndrome (trisomy 21)47, +21EitherLearning disability, typical facial features, heart defects
Turner syndrome45, XFemaleShort stature, ovaries do not work, usually infertile
Klinefelter syndrome47, XXYMaleSmall testes, usually infertile, sometimes breast growth
Triple X47, XXXFemaleOften no obvious signs; may be tall
XYY47, XYYMaleOften no obvious signs; may be tall
An embryo missing a whole autosome almost never survives to birth; 45, X is the only monosomy commonly seen in live births.
Practice this conceptself-check · 4 quick reps

Klinefelter is male, Turner is female

Sex follows the presence of the Y (with SRY), not the number of X chromosomes. XXY is male despite two X chromosomes; X with no second sex chromosome is female.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Linked genes and recombination frequency

    Recombination frequency

    RF=number of recombinant offspringtotal offspring×100%RF = \frac{\text{number of recombinant offspring}}{\text{total offspring}} \times 100\%
  • Building a gene map from crossover values

    Distances add along a chromosome

    d(A,B)=d(A,C)+d(C,B)when C lies between A and Bd(A,B) = d(A,C) + d(C,B) \quad \text{when C lies between A and B}
  • Meiosis and fertilisation as sources of genetic variation

    Chromosome combinations from independent assortment

    gametes=2nzygotes=2n×2n\text{gametes} = 2^{n} \qquad \text{zygotes} = 2^{n} \times 2^{n}

Reference tables (1)

Non-disjunction and aneuploidy: Down, Turner and Klinefelter syndromes5 rows
ConditionKaryotypeSexMain features
Down syndrome (trisomy 21)47, +21EitherLearning disability, typical facial features, heart defects
Turner syndrome45, XFemaleShort stature, ovaries do not work, usually infertile
Klinefelter syndrome47, XXYMaleSmall testes, usually infertile, sometimes breast growth
Triple X47, XXXFemaleOften no obvious signs; may be tall
XYY47, XYYMaleOften no obvious signs; may be tall
An embryo missing a whole autosome almost never survives to birth; 45, X is the only monosomy commonly seen in live births.

Watch out for (3)

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