Chromosome Mutations & Polyploidy

Linkage

Genes located near each other on the same chromosome TEND to be inherited together.

Why linked genes do not follow independent assortment

They do not behave as though they assort independently, because they are physically located near each other on the same chromosome.

Recombination

Crossing over can create chromosomes containing new combinations of parental alleles, i.e., a NOVEL combination of parental alleles.

Parental vs recombinant combinations: chromosome carrying A-B, homolog a-b

Parental combinations: AB and ab. Recombinant combinations: Ab and aB.

INPUT of a linkage testcross

A pair of homologous chromosomes carrying different allele combinations (a heterozygote), crossed with a tester (HOMOZYGOUS RECESSIVE).

Why use a homozygous recessive tester?

The tester only contributes recessive alleles, so the phenotype/genotype of the offspring tells you which gamete came from the heterozygous parent.

Do genes always assort independently?

No. Genes do not always assort independently (linkage is the exception).

Bateson & Punnett: the two traits studied

Flower color (P = purple, p = red) and pollen grain shape (L = long, l = round).

Bateson & Punnett: parental cross

PP LL × pp ll.

Bateson & Punnett: expected F1 genotype and phenotype

Pp Ll, with the phenotype purple and long.

If P/p and L/l assorted independently, what gametes would the F1 make?

Four gametes (PL, Pl, pL, pl) in equal proportion, a 1:1:1:1 ratio.

Expected F2 ratio for a regular dihybrid cross

9:3:3:1.

What does it mean when we DON'T see 9:3:3:1?

Linkage, which violates independent assortment, might be happening.

Excess parental phenotypes

Because of linkage there are a HUGE number of parental phenotypes (purple + long, red + round), occurring far more often than recombinant phenotypes. This violates the expected 9:3:3:1.

Key relationship in linked crosses

Parental phenotypes >>>> recombinant phenotypes.

Morgan's excess-parental-phenotype example: traits

Purple eye color (pr) and vestigial wings (vg) (in Drosophila).

Morgan's cross design

A heterozygous fly is testcrossed with a homozygous recessive.

Morgan's cross: which offspring classes are largest?

The two largest classes should be pr+/vg+ and pr/vg, which are the parental types.

Morgan's cross: what are the smaller classes?

The recombinants.

Recombinant frequency (RF) formula

RF = (recombinants / total offspring) × 100%.

Maximum expected RF if genes are completely unlinked

The maximum expected recombinant frequency approaches 50%.

Interpreting RF = 10%

If only 10% of offspring are recombinant, the genes are strongly linked.

Linked genes and the 50% benchmark

Linked genes produce fewer than 50% recombinants.

Why doesn't independent assortment apply to genes on the same chromosome?

They have physical linkage and tend to travel together during meiosis, so their alleles don't behave as though they are on completely separate chromosomes.

Mechanism that creates recombinants

Crossing over: during meiosis I, homologous chromosomes pair and non-sister chromatids exchange DNA.

Crossing-over example

Before: A-B and a-b. After crossing over: A-b and a-B, which are recombinant combinations.

Effect of linkage on parental combinations

Linkage causes parental combinations to be common.

Is crossing over reciprocal?

Yes. If one chromatid gives a piece to another chromatid, the exchange is reciprocal.

Where do chromatids cross over?

At the chiasmata (singular: chiasma).

Linkage relative to Mendel's second law

Linkage is an exception to independent assortment.

Chi-squared test for linkage: assumption 1

Normal segregation of alleles. Linkage would go against this.

Chi-squared test for linkage: assumption 2

Equal viability of genotypes. Some genotypes could survive better than others.

Degrees of freedom for a typical two-gene cross in this chi-squared test

(2 − 1)(2 − 1) = 1 degree of freedom.

Chi-squared formula

χ² = Σ (O − E)² / E. Asks: do observed data differ significantly from an expected distribution?

Map unit

1 map unit = 1% recombination (1 cM).

Genetic map distance

The frequency of recombination is the genetic map distance.

What do map distances predict?

The frequency of recombinants.

Important limitation of linkage map distances

Large distances become underestimated because of multiple crossovers.

Purpose of mapping studies

Associate phenotype WITH genotype; define the genetic interval of a trait for gene identification.

Crossover distribution: effect of physical distance

As the physical distance between genes increases, more crossovers occur.

Short vs long genetic intervals

A short interval has fewer recombinants; as the interval increases recombinant frequency increases BUT not linearly.

Why isn't RF linear with interval size?

Once genes are far apart, multiple crossovers can occur. Many restore the parental arrangement and become invisible when we simply count recombinant offspring.

As linkage increases and genetic distance decreases

Parental combinations >>> recombinants, and thus non-crossovers >>> crossovers (COs).

Can RF exceed 50%?

NO. Recombination frequency cannot exceed 50%.

If multiple crossovers occur, could RF exceed 50%?

NO. An individual crossover involves only two non-sister chromatids (not all 4), so it never results in RF over 50%.

Double crossover types

Two-strand, three-strand and four-strand double crossovers can produce different numbers of recombinant chromatids, but averaged across the possible products RF is still less than 50%.

What does 50% recombinants basically mean?

Independent assortment.

Map distances are additive: worked example

A—5 cM—B and A—3 cM—C, so C—2 cM—B.

What can recombination distances determine?

Gene order (recombination order can determine gene order).

What does a two-gene testcross tell you?

Whether genes are linked vs unlinked, and the approximate distance.

What does a three-gene (three-point) testcross tell you?

Gene order, distance between adjacent genes, distance across the whole region, double-crossover frequency, and interference.

Parents in the three-point cross (A B C)

Parents are always homozygous dominant (A B C) and homozygous recessive (a b c); the F1 heterozygote carries parental chromosomes A B C and a b c and is testcrossed.

Three-point cross: Region 1 and Region 2

Region 1 = interval A → B. Region 2 = interval B → C.

Three-point cross: crossover in Region 1 only (single crossover)

Just the A alleles swap relative to B and C: gives A b c and a B C.

Three-point cross: crossover in Region 2 only (single crossover)

The A and B alleles swap together relative to C (equivalently only C switches): gives A B c and a b C.

Three-point cross: crossovers in Region 1 AND 2 (double crossover, DCO)

The A and C alleles swap together relative to B (equivalently only the middle gene B switches): gives A b C and a B c.

All 8 offspring classes of the three-point testcross

Parentals: A B C, a b c. Region 1 SCO: A b c, a B C. Region 2 SCO: A B c, a b C. DCO: A b C, a B c.

How to get region distances from the 8 classes

Calculate the recombination frequency of each class group and work out the distance of each region.

Genetic interference

Crossovers in one region affect whether a crossover occurs in the neighboring region.

Expected DCO probability (worked example)

P(DCO) = (Region 1)(Region 2) = (0.18)(0.029) = 0.00522.

Expected number of DCOs (worked example)

0.00522 × 986 total offspring = 5.147, so about 5.1 DCOs expected; we may only observe 3, for instance.

Positive interference

Fewer DCOs observed than expected (E > O).

Three-point problem solving: what you are given

All genotype possibilities, including parental and recombinant classes.

Three-point problem solving: which classes to look at first

Specifically look at the parent classes and the DCO classes.

Why DCOs reveal the middle gene

A double crossover occurs once in Region I and once in Region II, so the allele gets switched relative to the two outside genes.

How to find gene order from parental and DCO classes

Compare each parental class with the corresponding DCO class: the gene that switches is the middle gene.

Worked example: parental A B C vs DCO A B c

Only c changes, so c is the middle gene. The lecture notes wrote the order as 'a b c or c b a'; if c is the middle gene the order should be a c b (or b c a), so check this against your problem set.

Once the order is known, how do you get interval lengths?

Calculate RF for each interval, remembering that DCOs are counted in BOTH intervals.

Calculating genetic interference: expected DCO

P(DCOs) = (region 1 length)(region 2 length) (× 100% if in percent); expected DCOs = P(DCOs) × total offspring.

Interference formula

I = (1 − O/E) × 100%, comparing observed (O) to expected (E) DCOs.

If E > O

Positive interference.

Coefficient of coincidence (CoC)

CoC = OBSERVED / EXPECTED DCOs.

CoC = 1

No interference.

CoC < 1

Positive interference.

CoC > 1

Negative interference.

Genetic linkage map vs physical map

Genetic linkage map: based on RECOMBINATION FREQUENCY (cM or map units). Physical map: based on NUMBER OF BASE PAIRS (bp, kb, Mb).

Average relationship between physical and genetic distance (as given)

1 Mb on average = 1 cM.

Linkage maps (type)

Based on recombination.

Cytological maps (type)

Based on chromosome banding and painting.

Physical maps (type)

Based on DNA sequence assembly.

Why use more than one map type?

Sometimes two maps are better than one.

Specialized mapping technique: centromere mapping

Can determine distances between genes and centromeres using recombination patterns.

Specialized mapping technique: molecular markers

DNA sequences that can be followed through families/populations.

Specialized mapping technique: linkage in human pedigrees

Track whether a marker and a trait are inherited together across generations.

Specialized mapping technique: mapping functions

Mathematical corrections for the fact that recombination frequency underestimates actual crossover frequency at larger distances.

Specialized mapping technique: physical maps

Use actual DNA/chromosome information.

Microsatellites

Researchers can use microsatellites as molecular markers; they can be traced through families and linked to a disease locus.

How do microsatellite markers indicate linkage?

If a particular marker pattern tracks with the phenotype across multiple family members, the marker is likely linked.

Why is human linkage analysis difficult?

Testcrosses are rare, progeny numbers are small, and genetic distances can be large.

Small sample size and RF

Small sample → uncertain RF estimate.

LOD score

Logarithm of odds: compares P(linkage) vs P(no linkage).

LOD formula

LOD = log10 [ P(linkage) / P(no linkage) ].

LOD ≥ +3

Strong evidence for linkage.

LOD ≤ −2

Strong evidence against linkage at that recombination fraction.

LOD = 0

The linkage model and no-linkage model are equally likely.

Chi-squared vs LOD: question each answers

Chi-squared: do observed data differ significantly from an expected distribution? LOD: is a linkage model more likely than a no-linkage model?

Why are map distances underestimated as distance increases?

Uncounted multiple crossovers: as the distance between two genes increases, more multiple crossovers occur and can go uncounted when only looking at the final recombination products.

Undetected multiple crossover example

A → b → A: the final chromosome looks parental again.

Effect on accuracy

Larger distance = less accurate recombination frequencies; observed map distance is underestimated as intervals get larger.

Why are mapping functions introduced?

They fix the underestimation problem by adjusting the observed proportion of recombinant genotypes to give a better estimate of actual genetic distance.