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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
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 offspring, n = 2, (p+q)2 = 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 offspring, n = 3, (p+q)3 = 1
(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 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 offspring, n = 2, (p+q)2 = 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 offspring, n = 3, (p+q)3 = 1
(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 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 offspring, n = 2, (p+q)2 = 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 offspring, n = 3, (p+q)3 = 1
(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
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
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
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
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
STATISTICS: BINOMIAL EXPANSION
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
STATISTICS: BINOMIAL EXPANSION
(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 ) =
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
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
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
genotypesP(II1 is A/a) = 1P(II2 is a/Y) = 1
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