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## LECTURE 03: PATTERNS OF INHERITANCE II

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LECTURE 03: PATTERNS OF INHERITANCE II

- genetic ratios & rules
- statistics
- binomial expansion
- Poisson distribution
- sex-linked inheritance
- cytoplasmic inheritance
- pedigree analysis

GENETIC RATIOS AND RULES

- product rule: the probability of independent events occurring together is the product of the probabilities of the individual events... AND

A/a x A/a

½ A + ½ a½ A + ½ a

P(a/a) = ½x½ = ¼

- sum rule: the probability of either of two mutually exclusive events occurring is the sumof the probabilities of the individual events... OR

A/a x A/a

½ A+ ½ a½ A+ ½ a P(A/a) = ¼+¼ = ½

STATISTICS: BINOMIAL EXPANSION

- diploid genetic data suited to analysis (2 alleles/gene)
- examples... coin flipping
- product and sum rules apply

n

- use formula: (p+q)n = [n!/(n-k)!k!] (pn-kqk) = 1

k=0

- define symbols...

STATISTICS: BINOMIAL EXPANSION

n

- use formula: (p+q)n = [n!/k!(n-k)!] (pkqn-k) = 1

k=0

- define symbols...

- p = probability of 1 outcome, e.g., P(heads)
- q = probability of the other outcome, e.g., P(tails)
- n = number of samples, e.g. coin tosses
- k = number of heads
- n-k= number of tails
- = sum probabilities of combinations in all orders

STATISTICS: BINOMIAL EXPANSION

n

- use formula: (p+q)n = [n!/k!(n-k)!] (pkqn-k) = 1

k=0

- e.g., outcomes from monohybrid cross...
- p = P(A_) = 3/4, q = P(aa) = 1/4
- k = #A_, n-k = #aa
- 2 possible outcomes if n = 1: k = 1, n-k = 0 ork = 0, n-k = 1
- 3 possible outcomes if n = 2: k = 2, n-k = 0 or k = 1, n-k = 1 ork = 0, n-k = 2
- 4 possible outcomes if n = 3... etc.

STATISTICS: BINOMIAL EXPANSION

n

- use formula: (p+q)n = [n!/k!(n-k)!] (pkqn-k) = 1

k=0

- e.g., outcomes from monohybrid cross...

STATISTICS: BINOMIAL EXPANSION

- Q: True breeding black and albino cats have a litter of all black kittens. If these kittens grow up and breed among themselves, what is the probability that at least two of three F2 kittens will be albino?
- A: First, define symbols and sort out basic genetics... one character – black > albino, one gene B > b ...

STATISTICS: BINOMIAL EXPANSION

- possible outcomes for three kittens...

STATISTICS: BINOMIAL EXPANSION

n

- expansion of (p+q)n = [n!/k!(n-k)!](pkqn-k) = 1

k=0

(3!/3!0!)(¾)3(¼)0 = x 27/64 = 27/64 = P(0 albinos)

(3!/2!1!)(¾)2(¼)1 = x 9/64 = 27/64 = P(1 albino)

(3!/1!2!)(¾)1(¼)2 = x 3/64 = 9/64 = P(2 albinos)

(3!/0!3!)(¾)0(¼)3 = x 1/64 = 1/64 = P(3 albinos)

P(at least 2 albinos) = 9/64 + 1/64 = 5/32

- ... if p(black) = ¾, q(albino) = ¼, n = 3 ...
- expansion of (p+q)3 = (¾+¼)(¾+¼)(¾+¼ ) =

}

CYTOPLASMIC INHERITANCE

selfing:

- are there differences between groups?
- are they true-breeding “genotypes”?

CYTOPLASMIC INHERITANCE

reciprocal crosses:

- are differences due to non-autosomal factors?
- compare progeny to see cytoplasmic influence

STATISTICS: POISSON DISTRIBUTION

- binomial... sample size (n) 10 or 15 at most
- if n = 103 or even 106 need to use Poisson
- e.g. if 1 out of 1000 are albinos [P(albino) = 0.001], and 100 individuals are drawn at random, what are the probabilities that there will be 3 albinos ?
- formula: P(k) = e-np(np)kk!

STATISTICS: POISSON DISTRIBUTION

- formula: P(k) = e-np(np)k k!
- k (the number of rare events)
- e= natural log = (1/1! + 1/2! + … 1/!) = 2.71828…
- n = 100
- p = P(albino) = 0.001
- np =P(albinos in population) = 0.1
- P(3) = e–0.1(0.1)3/3! = 2.718–0.1(0.001)/6 = 1.5 × 10–4

PEDIGREE ANALYSIS

- pedigree analysis is the starting point for all subsequent studies of genetic conditions in families
- main method of genetic study in human lineages
- at least eight types of single-gene inheritance can be analyzed in human pedigrees
- goals
- identify mode of inheritance of phenotype
- identify or predict genotypes and phenotypes of all individuals in the pedigree
- ... in addition to what ever the question asks

PEDIGREE ANALYSIS

- order of events for solving pedigrees
- identify all individuals according number and letter
- identify individuals according to phenotypes and genotypes where possible
- for I generation, determine probability of genotypes
- for I generation, determine probability of passing allele
- for II generation, determine probability of inheriting allele
- for II generation, same as 3
- for II generation, same as 4… etc to finish pedigree

PEDIGREE ANALYSIS

- additional rules...
- unaffected individuals mating into a pedigree are assumed to not be carriers
- always assume the most likely / simple explanation, unless you cannot solve the pedigree, then try the next most likely explanation

PEDIGREE ANALYSIS

Autosomal Recessive: Both sexes affected; unaffected parents can have affected progeny; two affected parents have only affected progeny; trait often skips generations.

PEDIGREE ANALYSIS

Autosomal Dominant: Both sexes affected; two unaffected parents cannot have affected progeny; trait does not skip generations.

PEDIGREE ANALYSIS

X-Linked Recessive: More affected than ; affected pass trait to all ; affected cannot pass trait to ; affected may be produced by normal carrier and normal .

PEDIGREE ANALYSIS

X-Linked Dominant: More affected than ; affected pass trait to all but not to ; unaffected parents cannot have affected progeny.

PEDIGREE ANALYSIS

Y-Linked: Affected pass trait to all ; not affected and not carriers.

Sex-Limited: Traits found in or in only.

Sex-Influenced, Dominant: More affected than ; all daughters of affected are affected; unaffected parents cannot have an affected .

Sex-Influenced, Dominant: More affected than ; all sons of an affected are affected; unaffected parents cannot have an affected .

PEDIGREE ANALYSIS

- mode of inheritance ?
- autosomal recessive
- autosomal dominant
- X-linked recessive
- X-linked dominant
- Y-linked
- sex limited

PEDIGREE ANALYSIS

- if X-linked recessive, what is the probability that III1 will be an affected ?

PEDIGREE ANALYSIS

A/Y a/a

A/a a/Y

A a a Y

a/Y

- genotypes

P(II1 is A/a) = 1

P(II2 is a/Y) = 1

- gametes

P(II1 passes A) = (1)(½)

P(II1 passes a) = (1)(½)

P(II2 passes a) = (1)(½)

P(II2 passes Y) = (1)(½)

PEDIGREE ANALYSIS

A/Y a/a

A/a a/Y

A a a Y

a/Y

- genotypes

P(II1 is A/a) = 1

P(II2 is a/Y) = 1

- gametes

P(II1 passes A) = (1)(½)

P(II1 passes a) = (1)(½)

P(II2 passes a) = (1)(½)

P(II2 passes Y) = (1)(½)

- P(III1 is affected ) = (1)(½) (1)(½) = ¼

PATTERNS OF INHERITANCE: PROBLEMS

- in Griffiths chapter 2, beginning on page 62, you should be able to do ALL of the questions
- begin with the solved problems on page 59 if you are having difficulty
- check out the CD, especially the pedigree problems
- try Schaum’s Outline questions in chapter 2, beginning on page 66, and chapter 5, beginning on page 158

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