Transcript Announcements

Announcements
1. Answers to Ch. 3 problems 6, 7, 8, 12, 17, 22, 32, 35
posted - 230A.
2. Problem set 1 answers due in lab this week at the
beginning of lab. Bring calculators to lab this week.
3. Getting to know Flylab and testcross (lab 2) - printout of
assignments from “notebook”, due this week at the
beginning of lab.
4. Confusion with X-linked crosses: 1 cross or 2?
5. Seminar this Thursday - faculty research interests
Review of last lecture
1. Genetic ratios are expressed as probabilities. Thus,
deriving outcomes of genetic crosses relies on an
understanding of laws of probability, in particular: the sum
law, product law, conditional probability (likelihood that one
outcome will occur, given a particular condition), and the
binomial theorum (used to determine particular
combinations). Expand the binomial OR use factorial
general formula to solve.
2. Statistical analyses (Chi square) - used to test the validity of
experimental outcomes. In genetics, some variation is
expected, due to chance deviation.
Outline of Lecture 6
I.
Chi-square revisited
II. Pedigree analysis- recessive vs. dominant traits
- solving pedigree problems
Chi-square formula
(o  e)
X 
e
2
2
where o = observed value for a given category,
e = expected value for a given category, and sigma is the
sum of the calculated values for each category of the ratio
• Once X2 is determined, it is converted to a probability
value (p) using the degrees of freedom (df) = n- 1
where n = the number of different categories for the
outcome.
Chi-square - Example 1
Phenotype
Expected
Observed
A
750
760
a
250
240
1000
1000
Null Hypothesis: Data fit a 3:1 ratio.
2
2
2










o

e
760

750
240

250
2
  



750
250
 e  

 2  0.53
degrees of freedom = (number of categories - 1) = 2 - 1 = 1
Use Fig. 3.12 to determine p - on next slide
X2 Table and Graph
Unlikely:
Reject hypothesis
likely
unlikely
Likely:
Do not reject
Hypothesis
0.50 > p > 0.20
Figure 3.12
Interpretation of p
• 0.05 is a commonly-accepted cut-off point.
• p > 0.05 means that the probability is greater than 5%
that the observed deviation is due to chance alone;
therefore the null hypothesis is not rejected.
• p < 0.05 means that the probability is less than 5%
that observed deviation is due to chance alone;
therefore null hypothesis is rejected. Reassess
assumptions, propose a new hypothesis.
Conclusions:
• X2 less than 3.84 means that we accept the Null
Hypothesis (3:1 ratio).
• In our example, p = 0.48 (p > 0.05) means that we
accept the Null Hypothesis (3:1 ratio).
• This means we expect the data to vary from
expectations this much or more 48% of the time.
Conversely, 52% of the repeats would show less
deviation as a result of chance than initially observed.
X2 Example 2: Coin Toss
I say that I have a non-trick coin (with both heads and
tails).
Do you believe me?
1 tail out of 1 toss
10 tails out of 10 tosses
100 tails out of 100 tosses
Tossing Coin - Which of these outcomes seem likely to you?
Compare Chi-square with 3.84 (since there is 1 degree of
freedom).
a) Tails
1 of 1
b) Tails
10 of 10
c) Tails
100 of 100
2
2
1
1
1 1 1 




1   0  
  
 2   2 
2 2 2 
2 

1
1
1
a)
2
2
Chi-square
b)
2 
2 
c)
10  52  0  52
5
 10
100  502  0  502
50
 100
Don’t reject
Reject
Reject
X2 - Example 3
F2 data: 792 long-winged (wildtype) flies, 208 dumpywinged flies.
Hypothesis: dumpy wing is inherited as a Mendelian
recessive trait.
Expected Ratio?
X2 analysis?
What do the data suggest about the dumpy mutation?
II. Pedigree analysis
The complex study of human genetics - we don’t control
human matings! Instead, we study family trees (pedigrees)
to identify how traits are inherited.
Importance of Pedigrees
• Genetic counselors use them to identify risk of
inherited illness.
• Genetic researchers use them to identify genes
responsible for genetic disorders.
• If you or a relative keep good family records, they
may some day be useful in tracking down a genetic
illness in your family.
Recessive vs. Dominant Traits
Autosomal Recessive Traits
• Example: The albino (aa) mutation inactivates the gene
for tyrosinase enzyme, which normally converts tyrosine
to melanin in the skin, hair and eyes.
• Non-albino is AA or Aa
• Autosomal recessive traits can skip generations (appear
in progeny of unaffected persons) and affect both males
and females equally.
Albinism Trait
Autosomal Dominant Traits
• Example: Hypotrichosis, hair loss begins in childhood
for both males and females.
• Autosomal dominant traits do not skip generations and
affect both males and females.
– Some but not all children will be affected in every
generation.
• Affected individuals are usually heterozygous since
mutant allele is rare.
Hypotrichosis Trait
Achondroplasia Trait (DD or Dd)
D is a dominant allele interfering with bone growth.
Most people are dd.
Human Recessive and Dominant Traits
OMIM: Online Mendelian Inheritance
in Man (Humans)
• Compiled by team at Johns Hopkins University.
• Available at: www.ncbi.nlm.nih.gov/omim/
• Includes description of trait, mode of inheritance,
molecular information.
• GREAT resource to find good reference for your
presentation topic; database is easy to search and full
of interesting information
• Another resource to find a topic from NPR
programming:
Symbols used in human pedigree analysis
Male
Female
Mating
Consanguineous marriage
Parents and 1 boy, 1 girl (in order of birth)
Sex unspecified
Affected individuals
Heterozygotes for autosomal recessive
Death
3
2
Number of children of sex indicated
Propositus
Pedigree Problem 1 - albinism
How is this trait inherited? dominant or recessive?
What are most probable genotypes of each individual?
Solving Pedigree Problems
• Inspect the pedigree:
– If trait is dominant, it will not skip generations nor be
passed on to offspring unless parents have it.
– If trait is recessive, it will skip generations and will
exist in carriers.
• Form a hypothesis, e.g. autosomal recessive.
• Deduce the genotypes.
• Check that genotypes are consistent with phenotypes.
• Revise hypothesis if necessary, e.g. autosomal
dominant.
What is the Mode of Inheritance?
Unaffected 2nd
generation =
rare, recessive
trait
Deduce genotypes
Pedigree Example 2: p. 71, #26
Pedigree Example 3:
Huntington’s Disease
unaffected
affected
Folksinger Woody Guthrie died of Huntington’s Disease.
Problem 3 Solution
HD = Huntington’s allele, hd = wildtype allele; data
suggests Huntington’s Disease is autosomal dominant.
Inconclusive Pedigree
• Try deducing the
genotypes using either
autosomal recessive or
autosomal dominant
hypothesis; either one
should work.