HMB265H1 Lecture Notes - Lecture 9: Null Hypothesis, Mendelian Inheritance, Genetic Linkage
Lecture 9: Linkage Mapping
• When two genes are on the same chromosome they are linked
• Recombinant progeny occurs when genes are linked
• The closer the genes are together on the chromosome the less likely
crossing over will occur and less chance of recombinant progeny
o Crossover frequency is a function of the distance between two
loci
• We can test whether two genes are linked by performing a test cross
and using the Chi-Squared Test
2 Chi-Squared Test
• Assesses the likelihood that a deviation from expectation is due to
chance
• Transmission of gametes is based on chance events
o Deviations from 1:1:1:1 ratio can represent chance events OR
linkage
o Ratios alone will never allow you to determine if observed data are significantly different from predicted
values
o The larger your sample, the closer your observed values are expected to match the predicted values
• Chi-Squared Test measures “goodness of fit” between observed and expected (predicted_ results
o Accounts for sample size, or the size of the experiemental population
• Framing a hypothesis:
o Null Hypothesis: observed values are not different from the expected values
▪ No linkage between the genes
o Alternative Hypothesis: observed values are different from the expected values
▪ Genes are linked
• Step 1
o Formulate null hypothesis: genes are not linked and predicts 1:1:1:1 ratio of phenotypes
• Step 2
o Compute
• Step 3
o Determine degrees of freedom (number of independent measurements)
• Step 4
o Consult 2 chart of critical values
A.H. Sturtevant
• Frequency of crossovers between two genes is a function of their distance apart on the chromosome; created the
first genetic (linkage) map
• One map unit (m.u.) = one centiMorgan (cM) = 1% recombination
o One map unit is the distance between 2 genes where 1 product of meiosis out of 100 is recombinant
• In any given cross the recombination frequency cannot be greater than 50% (=unlinked)
• Genetic map can have >50 map units, since it is cumulative across a chromosome
o Farther apart the genes are the closer their values are to 50%
• If two genes are not found on the same chromosome the maximum amount of recombinant types is 50%
o From independent assortment
• A value of 0 would be completely linked and progeny will
always be parental types
• A value of 50 could be independent assortment or linkage
• Map units are additive
o Must use closely spaced genes and add them
together
• Started with a dihybrid test cross
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HMB265H1 Full Course Notes
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