Presentation is loading. Please wait.

Presentation is loading. Please wait.

Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes Long aristae (appendages on head) Gray body Red eyes Normal wings Red.

Similar presentations


Presentation on theme: "Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes Long aristae (appendages on head) Gray body Red eyes Normal wings Red."— Presentation transcript:

1 Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes Long aristae (appendages on head) Gray body Red eyes Normal wings Red eyes Wild-type phenotypes II Y I X IV III 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

2 Linkage Mapping: Using Recombination Data Cross true breeding parents of different phenotypes Cross heterozygous F 1 organisms with pure-breeding recessives (like a TEST CROSS) Count recombinants (ones that look different from parental phenotype) Geneticists say that Linked genes exhibit a recombination frequency less than 50%.

3 A linkage map – Is the actual map of a chromosome based on recombination frequencies Recombination frequencies 9% 9.5% 17% bcn vg Chromosome 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 crossover would “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 two genetic loci is proportional to the distance separating the loci. The recombination frequencies used to construct a linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depicted in Figure The distances between genes are expressed as map units (centimorgans), with one map unit equivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data. Figure 15.7

4 male flies with white eyes female flies with red eyes (wild type) The F 1 generation all had red eyes F 2 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.b Evidence 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.

5 A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778- wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings What is the recombination frequency between these genes?

6 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 eyes 751- black body/purple eyes 49- gray body/purple eyes 45- black body/red-eyes What is the recombination frequency between these genes?

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

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


Download ppt "Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes Long aristae (appendages on head) Gray body Red eyes Normal wings Red."

Similar presentations


Ads by Google