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Lecture 9

HMB265H1 Lecture Notes - Lecture 9: Null Hypothesis, Mendelian Inheritance, Genetic Linkage

Human Biology
Course Code
Maria Papaconstantinou

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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
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
Transmission of gametes is based on chance events
o Deviations from 1:1:1:1 ratio can represent chance events OR
o Ratios alone will never allow you to determine if observed data are significantly different from predicted
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
Started with a dihybrid test cross
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