lecture 02: patterns of inheritance i fannouncements fkey concepts fmendelian analysis: e1 gene e2...
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LECTURE 02: PATTERNS OF INHERITANCE I
announcements key concepts Mendelian analysis:
1 gene 2 genes n genes
statistics chi-square
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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
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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
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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
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GARDEN PEA POLLENATION
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MENDELIAN ANALYSIS what characteristics to study study? Mendel observed 7 different characters
one at a time
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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
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MENDELIAN ANALYSIS: 1 GENE
crosses are reciprocal
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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
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MENDELIAN ANALYSIS: 1 GENE
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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
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MENDELIAN ANALYSIS: 1 GENE
random union
alleles segregate
1 allele, = probability
particulate genes
alleles phenotypes
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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
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MENDELIAN ANALYSIS: 1 GENE
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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
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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
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MENDELIAN ANALYSIS: TEST CROSS
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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
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MENDELIAN ANALYSIS: 2 GENES
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MENDELIAN ANALYSIS: 2 GENES
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MENDELIAN ANALYSIS: 2 GENES
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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
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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
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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
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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 !
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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
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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
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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
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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
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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
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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
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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
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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
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STATISTICS: CHI-SQUARE ANALYSIS
observe 4 phenotypes in roughly 9:3:3:1 ratio
generate a hypothesis what does it mean ? test hypothesis...
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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%)
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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
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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)
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READING & PRACTICE PROBLEMS
chapter 2 ... not done yet but, summary key terms solved problems 1, 2 questions 1 - 4, 9, 12, 13