Lecture Outline 9/1/05
• Mendelian genetics
– Monohybrid and dihybrid crosses
– Calculating offspring proportions in the F1 and F2
• Some rules of probability
• Goodness of fit tests
• Start Pedigree analysis
New Exam Dates
Wed Sept 28
Wed Oct 26
Tue Nov 22
4:30 PM
4:30 PM
(in class)
Check the course web site for homework problems
(due Sept 12)
Gregor Mendel
• He proposed two basic laws of genetics:
– Law of random segregation
• Alleles segregate in meiosis to produce haploid gametes
– Law of independent assortment
• Different traits are inherited independently
– He also figured out particulate inheritance, dominance, genotype
vs phenotype and more, although he didn’t use those words
• You can see a translation of his original paper at
www.mendelweb.org
Question:
Mice can be black (B-) or brown (bb).
What are the expected offspring of the
cross B/b x B/b?
1. Write down the probability for each kind of
gamete produced by a B/b parent.
2. Combine gametes from Mom and Dad to
produce offspring classes
1
First write down the
genotypes of the parents
Then write down the gametes
they can produce
Branch Diagrams
(fill this in on your own)
Fig 2.1Combine gametes to determine progeny
genotypes and phenotypes
Rules of probability
• Product rule
– The probability of both of two independent events is
the product of the probabilities
• Pr(A and B) = Pr(A) x Pr(B)
Determining genotypes of F2
black mice
• The F2 black mice contain a mixture of
genotypes-- how can you tell which are
which?
• Sum rule:
– The probability of either of two independent and
mutually exclusive events is the sum of the
probabilities
• Pr (A or B) = Pr(A) + Pr(B)
– Cross to make F3?
– Test cross?
• -> what do you expect from each?
Use product rule for “AND” questions
Use sum rule for “OR” questions
Independent Assortment
• Now consider pairs of characters.
• Mendel asserted that different traits assort
independently when gametes are produced.
(Only works if traits are on different chromosomes
or are far enough apart that crossing over occurs
each generation.)
Question
• Mendel crossed pure breeding green-round
peas with pure breeding yellow-wrinkled
peas.
– What types of progeny do you expect in the F1?
– What types of progeny do you expect in the F2?
2
Cross F1 to make F2
Cross parents to make F1
First write down the
genotypes of the parents
1. Write down the genotypes of
the parents
Then write down the gametes
they can produce
2. Then determine the GAMETES made by each parent
Combine all gametes to
determine progeny types
3. Finally combine
gametes to predict
offspring types
Fig 2.11
Fig 2.11
Now do the same thing with
Branch Diagrams
A dihybrid cross gives 9:3:3:1 ratio
R/r;Y/y x R/r;Y/y
R locus
Offspring can be:
1/4 RR
Or
1/4 Rr
Or
1/4 rR
Or
1/4 rr
Y locus
Offspring can be:
Combine probabilities to get the complete
offspring genotypes
1/4 YY
Or
1/4 Yy
Or
1/4 yY
Or
1/4 yy
1/4*1/4 =
1/16 RR YY
1/4*1/4 =
etc
1/16 RR Yy
Fig 2.10
Etc.
Solving more complex crosses
using branch diagrams
• AaBbccDd x aaBbCcdd
– > What is the probability of getting the
homozygous recessive (aabbccdd)?
Chi-square test
Example from a test cross: G/g x g/g
Phenotype
Observed #
Expected #
Green
30
25
Yellow
20
25
Total
50
50
– > Separate into single traits, then use the
product rule to combine independent traits
We expect equal numbers of green and yellow peas.
– > 1/2 * 1/4 *1/2 * 1/2 = 1/32
Are the observed data consistent with our
expectation? We need a measure of “lack of fit”
3
Chi Square
goodness of fit test
• Use for experiments where you COUNT something and
want to test the fit to a prediction.
Χ2 =SUM of
(observed - expected) 2
(expected)
Chi-square test
• X2 = Sum ((Obs - Exp)2 / Exp)
Phenotype
Green
Yellow
Total
Observed #
30
20
50
Expected #
25
25
50
(add up all of the squared deviations from the expected value)
X2
= 52 / 25 + -52 / 25
=2.0
Remember, aways use COUNTS, not percentages
2.0 < 3.84
Chi-square test
• What does p<0.05 mean?
– Less than 5% chance of seeing a deviation at
least that large, if the hypothesis is true.
• How do you determine the degrees of
freedom?
– Number of independent classes (n-1)
df = nclasses-1
there > 5% probability observing a discrepancy that large
Chi-square
• Use counts, not percentages
• Compare your value to critical values
– If your observed value is too large, the fit is
worse than expected by chance.
For Next Time:
• Read the rest of Chapter 2
• Do the first homework assignment (on web)
www.uvm.edu/~dstratto/bcor101
• Degrees of freedom
• Usually n-1, where n is the number of phenotypic
classes
4