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Linkage, and the Mapping of Genes on Chromosomes (chpt 5) Linkage and meiotic recombination
Genes linked together on the same chromosome usually assort together Linked genes may become separated through recombination
Mapping
The frequency with which genes become separated reflects the physical distance between them
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Crossing over leads to separation of linked genes during meiosis
A A a a
x
B B b b
Gametes Parental Recombinant
a B A B a b
A
b
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Linkage at a sex-linked gene
Deviation from 1:1:1:1 ratio of phenotypes for males
Fig. 5.2
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Parental vs Recombinant classes are defined by the allele combinations in the gametes from the Parental generation
Fig 5.3
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Fig. 5.4 Linkage in Autosomal genes
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Genetic recombination between car and Bar genes on the Drosophila X chromosome
Fig. 5.7
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Mapping Functions
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Observed versus Predicted Outcomes
Transmission of gametes is based on chance events which are subject to random variation. Chi square test measures "goodness of fit" between observed and expected (predicted) results Chi square test is used in genetics to compare experimental data with the predicted outcomes based a genetic hypothesis. Hypotheses can include no linkage (random assortment), gene interaction (recessive epistasis), etc.
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Applying the chi square test
Framing a hypothesis
Null hypothesis observed values are not different from the expected values
For linkage studies no linkage is null hypothesis Expect a 1:1:1:1 ratio of gametes.
Alternative hypothesis observed values are different from expected values
Expect significant deviation from 1:1:1:1 ratio. For linkage studies genes are linked.
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Chi Square Procedure
Obtain data (not percentages) for each phenotypic class Calculate the expected frequencies for each class based on null hypothesis
(number observed - number expected)2 (number expected) Compute degrees of freedom (df = number of classes-1) Use chi square value and df to determine p value from table of critical chi square values Evaluate the significance of p. If the p is smaller than 0.05, reject null hypothesis. If p is larger than 0.05, do not reject null hypothesis.
Calculate 2 =
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probability density function for the chisquare distribution
df = 3.0
from http://www.statlets.com/
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Applying the chi square test to a linkage study
AB / ab x ab / ab
Fig 5.5
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Chi Square Experiment 1 & 2
2 = (observed expected)2 number expected
2 = (31 25)2 + (19 25)2 25 25 2 = (62 50)2 + (38 50)2 50 50
Experiment 1
= 2.88
Experiment 2
= 5.76
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Chi square table of critical values
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Genetic Maps
A. H. Sturtevant Percentage of recombination, or recombination frequency (RF) reflects the physical distance separating genes two 1 RF = 1 map unit (or 1 centiMorgan)
Fig 5.8
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5.9
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Limitations of two point crosses
Difficult to determine gene order if two genes are close together Actual distances between genes do not always add up Pairwise crosses are time and labor consuming
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Three Point Crosses: A faster more accurate method to map genes
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Fig. 5.10
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Analyzing the results of a three point cross
Look at two genes at a time and compare to parental
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Fig. 5.11 a,b
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Double recombinants indicate order of three genes
Fi.g 5.11 a,d
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177 vg b distance 252 + 241 + 131 + 118 X 100 = 17.7 m.u. 4197 vg pr distance 252 + 241 + 13 + 9 4197 b pr distance 131 + 118 + 13 + 9 4197 X 100 = 12.3 m.u. X 100 = 6.4 m.u.
Correction for Double Crossovers
vg b distance 252+241+131+118+2(13 + 9) 4197 X 100 = 18.7 m.u.
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Interference: The number of double crossovers may be less than expected
Sometimes the number of observable double crossovers is less than expected if the two exchanges are independent
Occurrence of one crossover reduces likelihood that another crossover will occur in adjacent parts of the chromosome Chromosomal interference crossovers do not occur independently Interference is not uniform among chromosomes or even within a chromosome
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Measuring interference
Coefficient of coincidence = ratio between actual frequency of dco and expected frequency of dco Interference = 1 coefficient of coincidence
If interference = 0, observed and expected frequencies are equal If interference = 1, no double crossovers can occur
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Summary of three point cross analysis
Cross true breeding mutant with wild-type Backcross heterozygote to homozygous recessive Analyze Testcross Progeny (males if sex linked) Parental class most frequent Double crossovers least frequent Determine order of genes based on parentals and recombinants Determine genetic distance between each pair of recombinants Calculate coefficient of coincidence and interference
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Genes chained together by linkage relationships are known as linkage groups
Fig. 5.15
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5.1
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Human Xchromosome Map
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5-19
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5-21
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4-17
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Summary of linkage and recombination
Genes close together on the same chromosome are linked and do not segregate independently Linked genes lead to a larger number of parental class than expected in double heterozygotes Mechanism of recombination is crossing over between nonsister chromatids Chiasmata are the visible signs of crossing over The farther away genes are, the greater the opportunity for chiasmata to form. Recombination frequencies approximately reflect the physical distances between genes. Recombination frequencies between two genes vary from 0% to 50%
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