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.

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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
• 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 a½ 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 a½ 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 ) =

}

 are homogametic

 are heterogametic

 all white

 all red

 all red

red and white

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
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
• 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 
• 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

• genotypes

P(II1 is A/a) = 1

P(II2 is a/Y) = 1

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