Mutant phenotypesShort aristae

Black body

Cinnabareyes

Vestigialwings

Brown eyes

Long aristae(appendageson head)

Gray body

Redeyes

Normalwings

Redeyes

Wild-type phenotypes

IIY

IX IV

III

0 48.5 57.5 67.0 104.5

Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

Many fruit fly genes were mapped initially using recombination frequencies

T.H. Morgan did experiments with fruit flies to see how linkage affects the inheritance of two different characters ~ Linked genes that are close together on the same chromosome do not assort independently ~ Genes that assort independently are either:

- on separate chromosomes OR - are far apart on the same chromosome

Linkage Mapping: Using Recombination Data •Cross true breeding parents of different phenotypes

•Cross heterozygous F1 organisms with pure-breeding recessives (like a TEST CROSS)

• Count recombinants (ones that look different from parental phenotype)

•Geneticists say that Linked genesexhibit a recombination frequency less than 50%.

• A linkage map– Is the actual map of a chromosome based on

recombination frequencies

Recombinationfrequencies

9% 9.5%

17%

b cn vgChromosome

The b–vg recombination frequency is slightly less than the sum of the b–cn and cn–vg frequencies because double crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossoverwould “cancel out” the first and thus reduce the observed b–vg recombination frequency.

In this example, the observed recombination frequencies between three Drosophila gene pairs (b–cn 9%, cn–vg 9.5%, and b–vg 17%) best fit a linear order in which cn is positioned about halfway between the other two genes:

RESULTS

A linkage map shows the relative locations of genes along a chromosome.APPLICATION

TECHNIQUE A linkage map is based on the assumption that the probability of a crossover between twogenetic loci is proportional to the distance separating the loci. The recombination frequencies used to constructa linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depictedin Figure 15.6. The distances between genes are expressed as map units (centimorgans), with one map unitequivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data.

Figure 15.7

male flies with white eyesfemale flies with red eyes (wild type)

The F1 generation all had red eyes

F2 generation showed the 3:1 red:white eye ratio, but only males had white eyes

T.H. Morgan proposed that the white eye mutation was carried on X chromosome

Essential knowledge 3.A.3.bEvidence of student learning is a demonstrated understanding of each of the following: 3. The pattern of inheritance (monohybrid, dihybrid, sex-linked, and genes linked on the same homologous chromosome) can often be predicted from data that gives the parent genotype/ phenotype and/or the offspring phenotypes/genotypes.

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings.OFFSPRING: 778- wild type785- black-vestigial158- black- normal wings162- gray body-vestigial wings

What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for gray body and red eyes) is mated with a black fly with purple eyes.OFFSPRING: 721- gray body/red eyes751- black body/purple eyes49- gray body/purple eyes45- black body/red-eyes

What is the recombination frequency between these genes?

Determine the sequence of genes along a chromosome based on the following recombination frequencies:A-B = 8 map unitsA-C = 28 map unitsA-D = 25 map unitsB-C = 20 map unitsB-D = 33 map units

Determine the sequence of genes along a chromosome based on the following recombination frequencies:A-C = 20 map unitsA-D = 10 map unitsB-C = 15 map unitsB-D = 5 map units