presentation on gene linkage and mapping

1. Gene Linkage and Mapping
How to construct a Gene Map

2. Gene Linkage
 Law of Independent
Assortment
 Genes from different
chromosomes sort
independently in meiosis

3. What about genes on the same
chromosome?
 Thomas Morgan
 Linked genes- inherited together
 50 genes of the fruit fly
 4 linkage groups

4. Morgan’s Discovery
 Fruit fly has 4
chromosomes Gene
 Chromosomes
Chromosome
sort
independently,
genes do not
Image from Open Clipart Library

5. What about
Mendel and
his peas?

6. Gene Mapping
 Alfred Strutevant
 Hypothesized the rate at which crossing-over
separates linked genes could be used to map where
genes are located on a chromosome
 How to Construct a Gene Map
 Use the recombinant rate
 Shuffling of a genes into new combinations by
crossing-over in Prophase I of Meiosis

7. Gene Mapping Example
In fruit flies, the mutant black (b), has a body and the wild type has a grey
body; the mutant vestigial (v) has wings that are much shorter, compared
with the long wings of the wild type. In the following cross, true-breeding
parents are listed together with the counts of the offspring of F1 females
crossed with black and vestigial males:
P: black, normal x grey, vestigial
F1: females x black, vestigial
Progeny: 283 grey, normal
1,294 grey, vestigial
1,418 black, normal
241 black, vestigial
Use this data to calculate the map distances between black and vestigial
genes.

8. Gene Mapping Example continued
 1st step: identify the parental and recombinants
P: black, normal x grey, vestigial
F1: females x black, vestigial
Progeny: 283 grey, normal
1,294 grey, vestigial
1,418 black, normal
241 black, vestigial

9. Gene Mapping Example continued
P: black, normal x grey, vestigial
F1: females x black, vestigial
Progeny: 283 grey, normal
1,294 grey, vestigial
1,418 black, normal
241 black, vestigial
The parentals are the same phenotype as parents
 Grey, normal
 Black, vestigial

10. Gene Mapping Example continued
2nd step: Identify recombinants and figure
out percentage of progeny
P: black, normal x grey, vestigial
F1: females x black, vestigial
Progeny: 283 grey, normal
1,294 grey, vestigial
1,418 black, normal
241 black, vestigial

11. Gene Mapping Example continued
Recombinants :
grey, vestigial
black, normal
P: black, normal x grey, vestigial
F1: females x black, vestigial
Progeny: 283 grey, normal
1,294 grey, vestigial
1,418 black, normal
241 black, vestigial

12. Gene Mapping Example continued
To find percentage of recombination take each recombinant
and divide by the total offspring and multiply by 100.
P: black, normal x grey, vestigial
F1: females x black, vestigial
Progeny: 283 grey, normal
1,294 grey, vestigial
1,418 black, normal
241 black, vestigial

13. Gene Mapping Example continued
Progeny: 283 grey, normal
1,294 grey, vestigial
1,418 black, normal
241 black, vestigial
Total 3236
Recombinants :
grey, vestigial (1,294/3236) x 100 = 40%
black, normal (1,418/3235) x 100 = 43.8 %

14. Gene Mapping Example continued
Total recombination: 40% + 43.8% = 83.8%
Progeny: 283 grey, normal
1,294 grey, vestigial
1,418 black, normal
241 black, vestigial
Total 3236
Recombinants :
grey, vestigial (1,294/3236) x 100 = 40%
black, normal (1,418/3235) x 100 = 43.8 %

15. Gene Mapping Example continued
Change the percentage to map units
83.8% = 83.8 map units
The black and vestigial genes are located 83.8 map units apart.

16.  Images are from Chapters 11 and 15 of the Starr, Taggart,
Evers, and Starr’s Biology textbook.
 Starr, Cecie, Ralph Taggart, Christine Evers, and Lisa Starr.
Biology: The Unity and Diversity of Life. (2009). Brooks/Cole,
Cengage Learning.
 Image of chromosome from Open ClipArt Library user
gsagri04: http://openclipart.org/detail/134149/chromosome-
by-gsagri04