Lectures by Kathleen FitzpatrickSimon Fraser University

Copyright © 2012 Pearson Education Inc. Mark F. Sanders John L. Bowman

G E N E T I CA N I N T E G R A T E D A P P R O A C H

A N A LY S I S Chapter 5Genetic Linkage and

Mapping in Eukaryotes

CHAPTER 5, PART 1Genetics

Genetic Linkage and Mapping

• Thomas Hunt Morgan won the Nobel Prize for work establishing the chromosome theory of inheritance and also for his role identifying and explaining genetic linkage and recombination

• He applied linkage and recombination to genetic mapping

3

1:1:1:1 gamete ratio

What we have learned…All genes have been on different chromosomes

Allows “Independent Segregation”

However, gamete ratios can deviate from 1:1:1:1

Parental gametes

Meiosis I

A A a a

B B b b

A A a a

b b B B

or

AB or ab gametes Ab or aB gametes

Over all, 1 AB : 1 Ab : 1 aB : 1 ab

What will happen if two genes are on

the same chromosome?

Metaphase IAnaphase I

Meiosis – Two Genes on Same Chromosome

A A

B B

a a

b b

Usually only get AB or ab gamete (1:1 ratio)

Prophase I and Metaphase IAnaphase I

Will probably get a low frequency of Ab or aB

How?

“Parental-type gametes”

“Nonparental-type (recombinant-type) gametes”

Meiosis I; Prophase I

The Discovery of Genetic Linkage

• William Bateson and Reginald Punnett, in a series of crosses with sweet peas, found evidence of genetic linkage

• They crossed pure-breeding purple-flowered, long-pollen plants to white-flowered, round-pollen plants; the purple, long-pollen F1 were interbred to produce F2

• The expected 9:3:3:1 ratio was not observed

Discovery of Genetic Linkage

Pea flower color: Purple (P) is dominant to red (pp)

Pea pollen length: Long (L) is dominant to short (ll)

PPLL ppll Parents:

F1: All PpLl, (Purple, long)

F2: Purple, long:

Purple, short:

Red, long:

Red, short:

9

3

3

1

Expected Observed

284

21

21

55

Purple long Red short

How do you test if an observed ratio is significantly different from expected?

Chi Square Analysis

Class observed expected O-E

Purple, long

Purple, short

Red, long

Red, short

284

21

21

55

381

9/16 x 381

3/16 x 381

3/16 x 381

1/16 x 381

214.3

71.4

71.4

23.8

69.7

-50.4

-50.4

31.2

(O-E)2

4858

2540

2540

973

(O-E)2/ E

40.9

22.6

35.6

35.6

Σ = 134.7

df = n-1 = 4-1 = 3

Chi Square Probability Values

DF P= .99 .95 .90 .75 .50 .25 .10 .05 .01

3 0.114 0.351 0.584 1.212 2.365 4.108 6.251 7.814 11.34

P < .01c2

Conclusion?SIGNIFICANTLY DIFFERENT! Traits are not sorting independently!

Discovery of Linkage

Pea flower color: Purple (P) is dominant to red (pp)

Pea pollen length: Long (L) is dominant to short (ll)

PPLL ppll Parents:

F1: All PpLl, (Purple, long)

F2: Purple, long:

Purple, short:

Red, long:

Red, short:

Expected Observed

284

21

21

55

Significantly more!

Significantly more!

Significantly less!

Significantly less!

P and L genes appear to be “linked” together

215

71

71

24

Purple long Red short

What Bateson and Punnett Concluded

•Observation: MORE parental phenotypes, LESS nonparental types

• Conclusion: Bateson and Punnett suggested that an unknown mechanism kept the two parental gamete combinations together, which they called “coupling”

5.1 Linked Genes Do Not Assort Independently

• Genes located on the same chromosome are called syntenic genes

• Syntenic genes so close together that their alleles cannot assort independently are called linked genes

• Genetic linkage can be quantified to map the positions of genes on chromosomes

genemol.org

Recombination and Syntenic Genes

• Alleles of syntenic genes can be reshuffled when crossing over occurs between homologs to produce recombinant chromosomes

• Homologs that do not reshuffle alleles under study are called parental chromosomes or nonrecombinant chromosomes

• Genetic linkage mapping plots the positions of genes on chromosomes http://viirulentscience.wordpress.com/

Complete Genetic Linkage

• Complete genetic linkage is observed when no crossing over occurs between linked genes; only parental gametes are formed

• Some organisms exhibit complete linkage, e.g., Drosophila males have no crossing over

• The biological basis for this is unknown

Parental gametes

Incomplete Genetic Linkage

• Incomplete genetic linkage is much more common than complete linkage; in this case a mixture of parental and nonparental gametes are produced

Calculating Recombination Frequency

• Recombination frequency, expressed as r, is calculated as

• Recombination frequency is likely a reflection of the physical distance between two genes

r = # or recombinant animalstotal number of animals

Correlation Between Recombination Frequency and Gene Distance

• Crossing over occurs at a higher rate between genes that are farther apart, and a lower rate between genes that are closer together

Smaller r Bigger r

Morgan’s Crosses

• Morgan studied the white (eye color) and miniature (wing size) genes

• He crossed females pure-breeding for white eyes and small wings (wm/wm) to males that were wild type for both (wm/Y)

• The F1 were wm/wm females and wm/Y males

http://bioweb.wku.edu/courses/biol114/vfly1.asp

Morgan’s F1 F1 Results

• Morgan interbred the w+m+/wm females and wm/Y males

• A 1:1:1:1 ratio was predicted based on the assumption of independent assortment of the genes

• However, Morgan observed many more parental types than recombinant types, suggesting that the genes in question were found on the same chromosome, in this case the X

Less than 50%? →

Chi-Square Analysis of Morgan’s w, m Crosses

•

• There are 3 degrees of freedom in this case, and the p value is p 0.005

• Significant??

Yes! Thus the genes are linked!

Copyright © 2012 Pearson Education Inc. Genetics Analysis: An Integrated Approach

Interpretation of the Results

• Morgan suggested that nonparental allele combinations resulted from recombination between the X chromosomes of the heterozygous female parent

• He confirmed this explanation with many pairs of X-linked genes in Drosophila

Detecting Autosomal Genetic Linkage Through Test-Cross Analysis

• Morgan realized that linkage of autosomal genes in Drosophila could be interpreted using a two-point test-cross analysis

• In a test cross, the homozygous recessive parent contributes only recessive alleles

• Thus, the alleles contributed by just the dihybrid parent can be examined

Crosses with vg and pr

• Morgan crossed flies with purple eyes and vestigial (short) wings to wild type and obtained wild-type F1

• The F1 females were then crossed to males that had purple eyes and vestigial wings

• The alleles in the female gametes in this cross determined the phenotype in each of the progeny

• Offspring produced did not fit the 1:1:1:1 expected ratio

Important Conclusions from All of Morgan’s Crosses

1. Genetic linkage is a physical relationship between genes located near one another on a chromosome

2. Recombination occurs between linked genes less than 50% of the time, and greater than 50% of the gametes contain parental allele combinations

3. Recombination frequency varies among linked genes in proportion to the distance between them

5.2 Genetic Linkage Mapping Is Based on Recombination Frequency Between Genes

• Morgan recognized that with linked genes, more parental than recombinant progeny occurred and that the recombinant frequency varied among gene pairs

• Morgan suggested that closer proximity of genes produced a correspondingly higher frequency of parental allele combinations

• Therefore, we can use r values to make a map!

The First Genetic Linkage Map

• Morgan’s student, Alfred Sturtevant, realized that the variations in recombination frequency could be used to determine genetic maps for genes

• He used the results of several experiments to create a genetic map for five X-linked genes

http://www.ncbi.nlm.nih.gov

1. Of the genes tested, the pair with the smallest recombination frequency must be the closest in difference (y & w)

2. V is more distant from y than w, suggesting y-w-v

3. M is close to v, but more distant from w, so y-w-v-m

4. R is very far away from w, and fairly distant from v. This suggests that r is on the opposite end of the map, so y-w-v-m-r

General logic for generating the map:

You will try this in your homework!

Map Units

• Recombination frequencies between two genes can be converted into units of physical distance, map units (m.u.)

• A map unit is also called a centiMorgan (cM)

• By convention, 1% recombination 1 m.u. or 1 cM

http://www.tutorvista.com

Example Mapping Problem

A homozygous pea plant with purple flowers and long pollen (PPLL) is crossed with a second inbred line with red flowers and short pollen (ppll)

How would you show their genotypes using new method?PL

PL

pl

pl

What would the F1 look like?

PL

pl

Example Mapping Problem, Cont’

An F1 plant is testcrossed and the following progeny were observed:

# of Progeny

Purple, long 39

Purple, short 9

Red, long 10

Red, short 42

Calculate the map distance between the P and L genes.

PL

pl

Step 1: Write genotypes of parents

pl

pl

Step 2: Write genotypes of kids

Example Mapping Problem, Cont’

PL

pl

pl

pl

# of Progeny

Purple, long 39

Purple, short 9

Red, long 10

Red, short 42

From mom? From dad?pl

pl

pl

pl

PL

Pl

pL

pl

Which are the result of a parental-type gamete?Which are the result of a recombinant-type gamete?

Note: We are only looking at recombination in the heterozygous parent!

Example Mapping Problem, Cont’

# of Progeny

Purple, long 39

Purple, short 9

Red, long 10

Red, short 42

These 19 progeny were the result of a recombination between the “P” and “L” genes.

# RecombinantsTotal # of progeny

X 100Recombination % =

= [(9 + 10) / 100] x 100 = 19%

The P and L genes are 19 cM apart

What is the map distance?

Final Step: Draw the Map

You can then draw a map showing the distance between the two genes

19 cM

P L

This is Two Point Linkage Analyses

Practice Problems!

• Chapter 5: 2 & 3

Questions?