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

GENETIC RATIOS AND RULES Fsum 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/a  ½ A + ½ a  ½ A + ½ a P( A/a ) = ¼ + ¼ = ½ Fproduct 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 ) = ½ x ½ = ¼

STATISTICS: BINOMIAL EXPANSION F diploid genetic data suited to analysis (2 alleles/gene) F examples... coin flipping F product and sum rules apply n F use formula: ( p + q ) n =  [ n !/( n-k )! k !] ( p n-k q k ) = 1 k =0 F define symbols...

STATISTICS: BINOMIAL EXPANSION n F use formula: ( p + q ) n =  [ n!/k ! (n-k )!] ( p k q n-k ) = 1 k =0 F p = probability of 1 outcome, e.g., P (heads) F q = probability of the other outcome, e.g., P (tails) F n = number of samples, e.g. coin tosses F k = number of heads F n-k = number of tails F  = sum probabilities of combinations in all orders F define symbols...

STATISTICS: BINOMIAL EXPANSION n F use formula: ( p + q ) n =  [ n!/k ! (n-k )!] ( p k q n-k ) = 1 k =0 F e.g., outcomes from monohybrid cross... F p = P ( A_ ) = 3/4, q = P ( aa ) = 1/4 F k = # A_, n-k = # aa F 2 possible outcomes if n = 1: k = 1, n-k = 0 or k = 0, n-k = 1 F 3 possible outcomes if n = 2: k = 2, n-k = 0 or k = 1, n-k = 1 or k = 0, n-k = 2 F 4 possible outcomes if n = 3... etc.

STATISTICS: BINOMIAL EXPANSION n F use formula: ( p + q ) n =  [ n!/k ! (n-k )!] ( p k q n-k ) = 1 k =0 F e.g., outcomes from monohybrid cross...

STATISTICS: BINOMIAL EXPANSION F 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 F 2 kittens will be albino? F A: First, define symbols and sort out basic genetics... one character – black > albino, one gene B > b...

STATISTICS: BINOMIAL EXPANSION

F possible outcomes for three kittens...

STATISTICS: PASCAL’S TRIANGLE

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 F expansion of ( p + q ) n =  [ n!/k ! (n-k )!] ( p k q n-k ) = 1 k =0 } F... if p (black) = ¾, q (albino) = ¼, n = 3... F 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  red and white  all red

  selfing: H are there differences between groups? H are they true-breeding “genotypes”? CYTOPLASMIC INHERITANCE

  reciprocal crosses: H are differences due to non-autosomal factors? H compare  progeny to see cytoplasmic influence 

STATISTICS: POISSON DISTRIBUTION F binomial... sample size ( n )  10 or 15 at most F if n = 10 3 or even 10 6 need to use Poisson F 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 ? Fformula: P ( k ) = e -np ( np ) k k!

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

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

PEDIGREE ANALYSIS: SYMBOLS

PEDIGREE ANALYSIS F 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 F 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 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 F mode of inheritance ? F autosomal recessive  F autosomal dominant F X -linked recessive F X -linked dominant F Y -linked  F sex limited 

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

PEDIGREE ANALYSIS F genotypes P (II1 is A/a ) = 1 P (II2 is a/Y ) = 1 A/Y a/a A/a a/Y

PEDIGREE ANALYSIS F genotypes P (II1 is A/a ) = 1 P (II2 is a/Y ) = 1 F 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) A/Y a/a A/a a/Y A a a Y a/Y

PEDIGREE ANALYSIS F genotypes P (II1 is A/a ) = 1 P (II2 is a/Y ) = 1 F 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) F 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 F in Griffiths chapter 2, beginning on page 62, you should be able to do ALL of the questions F begin with the solved problems on page 59 if you are having difficulty F check out the CD, especially the pedigree problems F try Schaum’s Outline questions in chapter 2, beginning on page 66, and chapter 5, beginning on page 158