LECTURE 03: PATTERNS OF INHERITANCE II Fgenetic ratios & rules Fstatistics Ebinomial expansion EPoisson distribution Fsex-linked inheritance Fcytoplasmic.

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

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

A/a x A/a½ A + ½ a ½ A + ½ a

P(A/a) = ¼ + ¼ = ½

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

A/a x A/a½ A + ½ a ½ A + ½ a

P(a/a) = ½ x ½ = ¼

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

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

define symbols...

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 or k = 0, n-k = 1 3 possible outcomes if n = 2: k = 2, n-k = 0 or k = 1, n-k = 1 or k = 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...

1 offspring, n = 1, (p+q)1 = 1

(1!/1!0!)(¾)1(¼)0 = 3/4 (1!/0!1!)(¾)0(¼)1 = 1/4 = 1


(2!/2!0!)(¾)2(¼)0 = (1)9/16 (2!/1!1!)(¾)1(¼)1 = (2)3/16 (2!/0!2!)(¾)0(¼)2 = (1)1/16 = 1 16/16


(3!/3!0!)(¾)3(¼)0 = (1)27/64 (3!/2!1!)(¾)2(¼)1 = (3)9/64 P of 2A_ + 1aa (3!/1!2!)(¾)1(¼)2 = (3)3/64 (3!/0!3!)(¾)0(¼)3 = (1)1/16 = 1


(1!/1!0!)(¾)1(¼)0 = 3/4 (1!/0!1!)(¾)0(¼)1 = 1/4 = 1


(2!/2!0!)(¾)2(¼)0 = (1)9/16 (2!/1!1!)(¾)1(¼)1 = (2)3/16 (2!/0!2!)(¾)0(¼)2 = (1)1/16 = 1 16/16


(3!/3!0!)(¾)3(¼)0 = (1)27/64 (3!/2!1!)(¾)2(¼)1 = (3)9/64 P of 2A_ + 1aa (3!/1!2!)(¾)1(¼)2 = (3)3/64 (3!/0!3!)(¾)0(¼)3 = (1)1/16 = 1


(1!/1!0!)(¾)1(¼)0 = 3/4 (1!/0!1!)(¾)0(¼)1 = 1/4 = 1


(2!/2!0!)(¾)2(¼)0 = (1)9/16 (2!/1!1!)(¾)1(¼)1 = (2)3/16 (2!/0!2!)(¾)0(¼)2 = (1)1/16 = 1 16/16


(3!/3!0!)(¾)3(¼)0 = (1)27/64 (3!/2!1!)(¾)2(¼)1 = (3)9/64 P of 2A_ + 1aa (3!/1!2!)(¾)1(¼)2 = (3)3/64 (3!/0!3!)(¾)0(¼)3 = (1)1/16 = 1


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 ...


P1: black × albino genotype: BB bb

gametes: P(B) = 1 P(b) = 1

F1: black × black genotype: Bb Bb gametes: P(B) = ½ , P(b) = ½ P(B) = ½ , P(b) = ½

expected F2: P(black) = P(B_) = (½ )2 + 2(½ )2 = ¾ = p

P(albino) = P(bb) = (½ )2 = ¼ = q


gametes: P(B) = 1 P(b) = 1

F1: black × black genotype: Bb Bb gametes: P(B) = ½ , P(b) = ½ P(B) = ½ , P(b) = ½


P(albino) = P(bb) = (½ )2 = ¼ = q


gametes: P(B) = 1 P(b) = 1 F1: black × black

genotype: Bb Bb gametes: P(B) = ½ , P(b) = ½ P(B) = ½ , P(b) = ½


P(albino) = P(bb) = (½ )2 = ¼ = q


F2 * Probability 3 black + 0 albino P ( ) = (¾ )3 × (¼ )0 = 27/64 P ( ) = (¾ )2 × (¼ )1 = 9/64

P ( ) = (¾ )2 × (¼ )1 = 9/64 2 black + 1 albino P ( ) = (¾ )2 × (¼ )1 = 9/64

P ( ) = (¾ )1 × (¼ )2 = 3/64

P ( ) = (¾ )1 × (¼ )2 = 3/64 1 black + 2 albino P ( ) = (¾ )1 × (¼ )2 = 3/64

0 black + 3 albino P ( ) = (¾ )0 + (¼ )3 = 1/64 Total 64/64

possible outcomes for three kittens...

STATISTICS: PASCAL’S TRIANGLESample Frequency Possible Outcomes* Combinations

1 × 1 + 1 2 orders

2 × 1 + 2 + 1 3 orders

3 × +++ 4 orders

4 × 1 + 4 + 6 + 4 + 1 5 orders

5 × 1 + 5 + 10 + 10 + 5 + 1 6 orders

6 × 1 + 6 + 15 + 20 + 15 + 6 + 1 7 orders


(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

n expansion of (p+q)n = [n!/k!(n-k)!] (pkqn-k) = 1 k=0

}

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

SEX-LINKED INHERITANCE


are homogametic are heterogametic





all white all red


the white gene is X-linked

red and white all red

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

CYTOPLASMIC INHERITANCE

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)k k!

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: SYMBOLS

PEDIGREE ANALYSIS

order of events for solving pedigrees 1. identify all individuals according number and letter 2. identify individuals according to phenotypes and

genotypes where possible 3. for I generation, determine probability of genotypes 4. for I generation, determine probability of passing allele 5. for II generation, determine probability of inheriting

allele 6. for II generation, same as 3 7. for II generation, same as 4… etc to finish pedigree

PEDIGREE ANALYSIS

additional rules... 1. unaffected individuals mating into a pedigree are

assumed to not be carriers 2. always assume the most likely / simple explanation,

unless you cannot solve the pedigree, then try the next most likely explanation

PEDIGREE ANALYSISAutosomal Recessive: Both sexes affected; unaffected

parents can have affected progeny; two affected parents have only affected progeny; trait often skips generations.

PEDIGREE ANALYSISAutosomal Dominant: Both sexes affected; two

unaffected parents cannot have affected progeny; trait does not skip generations.

PEDIGREE ANALYSISX-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 ANALYSISX-Linked Dominant: More affected than ; affected

pass trait to all but not to ; unaffected parents cannot have affected progeny.

PEDIGREE ANALYSISY-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 .

I

II

III

PEDIGREE ANALYSIS

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

I

II

III

PEDIGREE ANALYSIS

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

I

II

III

PEDIGREE ANALYSIS

genotypesP(II1 is A/a) = 1P(II2 is a/Y) = 1

A/Y a/a

A/a a/Y

I

II

III

PEDIGREE ANALYSIS


gametesP(II1 passes A) = (1)(½)P(II1 passes a) = (1)(½)P(II2 passes a) = (1)(½)P(II2 passes Y) = (1)(½)

A/Y a/a

A/a a/Y

A a a Y

a/Y

I

II

III

PEDIGREE ANALYSIS


gametesP(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

A/Y a/a

A/a a/Y

A a a Y

a/Y

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

LECTURE 03: PATTERNS OF INHERITANCE II Fgenetic ratios & rules Fstatistics Ebinomial expansion EPoisson distribution Fsex-linked inheritance Fcytoplasmic.

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