LECTURE 02: PATTERNS OF INHERITANCE I

announcements key concepts Mendelian analysis:

1 gene 2 genes n genes

statistics chi-square

CHAPTER 2: KEY CONCEPTS(most of which I assume you already know)

existence of genes inferred by observing standard progeny ratios derived from controlled matings

discrete phenotypes can have single gene basis in a diploid cell, each gene is represented twice,

one allele on each chromosome pair

CHAPTER 2: MORE KEY CONCEPTS(most of which I assume you already know)

inheritance patterns based on behavior of chromosomes during meiosis

alleles of a gene segregate into different gametes gene pairs on different chromosomes assort

independently genes on sex chromosomes show unique

inheritance patterns

MENDELIAN ANALYSIS

what organism to study? Mendel used garden pea (Pisum sativum) available cheap from local merchants easy to grow, small space, fast generation time lots of phenotypic variation could be manipulated for controlled pollenation:

self and outcross

GARDEN PEA POLLENATION

MENDELIAN ANALYSIS what characteristics to study study? Mendel observed 7 different characters

one at a time

MENDELIAN ANALYSIS: 1 GENE

e.g.: purple vs white flowers must start with true-breeding or pure-breeding lines how do you know? selfing 1 phenotype nomenclature:

distinguish genotype & phenotype individuals are homozygotes or heterozygotes generations designated P, F1, F2 ... determine genotypic & phenotypic ratios

MENDELIAN ANALYSIS: 1 GENE

crosses are reciprocal

MENDELIAN ANALYSIS: 1 GENE

P

F1

F2

purple x white

purple x purple

¾ purple + ¼ white

white x purple

purple x purple

¾ purple + ¼ white

not blending because we see a return of white

MENDELIAN ANALYSIS: 1 GENE

MENDELIAN ANALYSIS: 1 GENE

1. particulate hereditary determinants (genes)2. alleles different phenotypes; dominance3. alleles of a gene segregate into different gametes4. gametes receive 1 allele with equal probability5. union of gametes at fertilization is random

MENDELIAN ANALYSIS: 1 GENE

random union

alleles segregate

1 allele, = probability

particulate genes

alleles phenotypes

MENDELIAN ANALYSIS: 1 GENE

Pgametes

F1

gametes

F2

ratio

purple whiteP/P x p/p

P pall purple

P/p x P/p½ P + ½ p ½ P + ½ p ¾ purple ¼ white¼ P/P + ½ P/p + ¼ p/p

1 : 2 : 1

MENDELIAN ANALYSIS: 1 GENE

MENDELIAN ANALYSIS: CROSSES

1 gene: e.g., F1 yellow heterozygotes

Y/y x Y/y monohybrid cross

1 gene: e.g., yellow unknown x green homozygoteY/? x y/y test cross

MENDELIAN ANALYSIS: CROSSES

1 gene: e.g., yellow heterozygote x green homozygoteY/y x y/y ½ Y/y & ½ y/y

or

yellow homozygote x green homozygoteY/Y x y/y all Y/y

1 gene: e.g., F1 yellow heterozygotes

Y/y x Y/y monohybrid cross

MENDELIAN ANALYSIS: TEST CROSS

MENDELIAN ANALYSIS: CROSSES

2 genes: e.g., F1 round yellow heterozygotes

R/r Y/y x R/r Y/y dihybrid cross

1 gene: e.g., F1 yellow heterozygotes

Y/y x Y/y monohybrid cross

1 gene: e.g., yellow unknown x green homozygoteY/? x y/y test cross

MENDELIAN ANALYSIS: 2 GENES

MENDELIAN ANALYSIS: 2 GENES

MENDELIAN ANALYSIS: 2 GENES

MENDELIAN ANALYSIS: 2 GENES

1. particulate hereditary determinants (genes)2. alleles different phenotypes; dominance3. alleles of a gene segregate into different gametes4. gametes receive 1 allele with equal probability5. union of gametes at fertilization is random6. alleles of different genes* assort independently

into different gametes * gene pairs on different chromosomes

MENDELIAN ANALYSIS: 2 LAWS

1. particulate hereditary determinants (genes)2. alleles different phenotypes; dominance3. alleles of a gene segregate into different gametes4. gametes receive 1 allele with equal probability5. union of gametes at fertilization is random6. alleles of different genes* assort independently into

different gametes * = gene pairs on different chromosomes

MENDELIAN ANALYSIS: 1 n GENES

3 methods of working out expected outcomes of controlled breeding experiments:1. Punnet square2. tree method (long) – genotypes3. tree method (short) – phenotypes

MENDELIAN ANALYSIS: PUNNET SQUARE

all pairings of and gametes = probabilities of all pairings some pairings occur >1 different Ps for different

genotypes & phenotypes 1 gene, 2 x 2 = 4 cells 2 genes, 4 x 4 = 16 cells 3 genes, 8 x 8 = 64 cells...

too much work !

MENDELIAN ANALYSIS: LONG TREE1 gene, alleles A, a

1/4 A/A

1/2 A/a

1/4 a/a

GENOTYPE

1/4 A/A

1/2 A/a

1/4 a/a

MENDELIAN ANALYSIS: LONG TREE1 gene, alleles A, a

1/4 A/A

1/2 A/a

1/4 a/a

GENOTYPE

1/4 A/A

1/2 A/a

1/4 a/a

PHENOTYPE

3/4 A

1/4 a

MENDELIAN ANALYSIS: LONG TREE2 genes, alleles A, a; B, b...

1/4 B/B 1/4 A/A 1/2 B/b

1/4 b/b 1/4 B/B

1/2 A/a 1/2 B/b 1/4 b/b 1/4 B/B

1/4 a/a 1/2 B/b 1/4 b/b

GENOTYPE

1/16 A/A B/B1/8 A/A B/b1/16 A/A b/b1/8 A/a B/B1/4 A/a B/b1/8 A/a b/b1/16 a/a B/B1/8 a/a B/b1/16 a/a b/b

MENDELIAN ANALYSIS: LONG TREE2 genes, alleles A, a; B, b...

1/4 B/B 1/4 A/A 1/2 B/b

1/4 b/b 1/4 B/B

1/2 A/a 1/2 B/b 1/4 b/b 1/4 B/B

1/4 a/a 1/2 B/b 1/4 b/b

GENOTYPE

1/16 A/A B/B1/8 A/A B/b1/16 A/A b/b1/8 A/a B/B1/4 A/a B/b1/8 A/a b/b1/16 a/a B/B1/8 a/a B/b1/16 a/a b/b

PHENOTYPE

9/16 AB

3/16 Ab

3/16 aB

1/16 ab

MENDELIAN ANALYSIS: SHORT TREE2 genes, alleles A, a; B, b...

¾ B ¾ A

¼ b

¾ B ¼ a

¼ b

much easier... can extend to >2 genes

PHENOTYPE

9/16 AB

3/16 Ab

3/16 Ab

1/16 ab

formulae for n genes, dominance, n hybrid crosses …

MENDELIAN ANALYSIS: 1 n GENES

1. # possible different types of gametes = 2n

2. # possible different genotypes of progeny = 3n

3. frequency of least common genotype = (1/4)n

4. # possible different phenotypes of progeny = 2n

STUFF TO THINK ABOUT

Mendel’s choice of characters was critical did he chose 7 characters that appeared to reside on

different chromosomes by chance? what would happen if some were linked? all exhibited dominant/recessive relationships

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: CHI-SQUARE ANALYSIS

observe 4 phenotypes in roughly 9:3:3:1 ratio

generate a hypothesis what does it mean ? test hypothesis...

STATISTICS: CHI-SQUARE ANALYSIS

O : E O-E (O-E)2 (O-E)2/E YR 315 9 313 2 4 0.013 Yr 108 3 104 4 16 0.154 yR 101 3 104 -3 9 0.087 yr 32 1 35 -3 9 0.257 556 16 556 2c = 0.511

go to table 2-2 on page 41... 4 groups... 3 degrees of freedom (df = 3) chose the probability that you are willing to

accept for mistakenly rejecting a hypothesis that is in fact true... ( = 0.05 or 5%)

STATISTICS: CHI-SQUARE ANALYSIS

O : E O-E (O-E)2 (O-E)2/E YR 315 9 313 2 4 0.013 Yr 108 3 104 4 16 0.154 yR 101 3 104 -3 9 0.087 yr 32 1 35 -3 9 0.257 556 16 556 2c = 0.511

P .995 .99 .975 .95 .9 .75 .5 0.25 .1 .05 .01 .001

df

1 .000 .000 .001 .004 .016 .102 .455 1.32 2.71 3.84 6.63 10.8 2 .010 .020 .051 .103 .211 .575 1.39 2.77 4.61 5.99 9.21 13.8 3 .072 .115 .216 .352 >2c> .584 1.21 2.37 4.11 6.25 7.81 11.3 16.3 4 .207 .297 .484 .711 1.06 1.92 3.36 5.39 7.78 9.49 13.3 18.5 5 .412 .554 .831 1.15 1.61 2.68 4.35 6.61 9.24 11.1 15.1 20.5

STATISTICS: CHI-SQUARE ANALYSIS

express this as: 0.95 > P(2c = 0.511) > 0.90

the data do not deviate significantly from a 9:3:3:1 ratio we do not reject our H0 (alternatively we could reject) “seed color and seed shape fit an unlinked, 2 gene

classic Mendelian model with complete dominance” the probability of deviation by chance from this model

lies between 90 and 95% (i.e., the biological explanation is supported by data)

READING & PRACTICE PROBLEMS

chapter 2 ... not done yet but, summary key terms solved problems 1, 2 questions 1 - 4, 9, 12, 13