genetics chapter 5 part 1
TRANSCRIPT
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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
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CHAPTER 5, PART 1Genetics
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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
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1:1:1:1 gamete ratio
What we have learned…All genes have been on different chromosomes
Allows “Independent Segregation”
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However, gamete ratios can deviate from 1:1:1:1
Parental gametes
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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
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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”
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Meiosis I; Prophase I
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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
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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?
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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!
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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
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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”
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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
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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/
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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
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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
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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
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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
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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
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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%? →
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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!
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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
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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
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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
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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
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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!
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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
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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!
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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
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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
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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
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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!
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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?
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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
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Practice Problems!
• Chapter 5: 2 & 3
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Questions?