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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 • 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 ) = }
SEX-LINKED INHERITANCE are homogametic are heterogametic
SEX-LINKED INHERITANCE all white all red
SEX-LINKED INHERITANCE the white gene is X-linked 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 • ... 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 • 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