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.
2.
3.
4.
5.
particulate hereditary determinants (genes)
alleles  different phenotypes; dominance
alleles of a gene segregate into different gametes
gametes receive 1 allele with equal probability
union of gametes at fertilization is random
MENDELIAN ANALYSIS: 1 GENE
particulate genes
alleles  phenotypes
alleles segregate
1 allele, = probability
random union
MENDELIAN ANALYSIS: 1 GENE
P
gametes
F1
gametes
F2
ratio
purple white
P/P x p/p
P  p
all purple
P/p x P/p
½P+½p  ½P+½p
¾ purple ¼ white
¼ 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 homozygote
Y/? x y/y  test cross
MENDELIAN ANALYSIS: CROSSES
1 gene: e.g., F1 yellow heterozygotes
Y/y x Y/y  monohybrid cross
1 gene: e.g., yellow heterozygote x green homozygote
Y/y x y/y  ½ Y/y & ½ y/y
or
yellow homozygote x green homozygote
Y/Y x y/y  all Y/y
MENDELIAN ANALYSIS: TEST CROSS
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 homozygote
Y/? x y/y  test cross
2 genes: e.g., F1 round yellow heterozygotes
R/r Y/y x R/r Y/y  dihybrid cross
MENDELIAN ANALYSIS: 2 GENES
MENDELIAN ANALYSIS: 2 GENES
MENDELIAN ANALYSIS: 2 GENES
MENDELIAN ANALYSIS: 2 GENES
1.
2.
3.
4.
5.
6.
particulate hereditary determinants (genes)
alleles  different phenotypes; dominance
alleles of a gene segregate into different gametes
gametes receive 1 allele with equal probability
union of gametes at fertilization is random
alleles of different genes* assort independently
into different gametes
* gene pairs on different chromosomes
MENDELIAN ANALYSIS: 2 LAWS
1.
2.
3.
4.
5.
6.
particulate hereditary determinants (genes)
alleles  different phenotypes; dominance
alleles of a gene segregate into different gametes
gametes receive 1 allele with equal probability
union of gametes at fertilization is random
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 square
2. tree method (long) – genotypes
3. 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 TREE
1 gene, alleles A, a
GENOTYPE
1/4 A/A

1/4 A/A
1/2 A/a

1/2 A/a
1/4 a/a

1/4 a/a
MENDELIAN ANALYSIS: LONG TREE
1 gene, alleles A, a
GENOTYPE
PHENOTYPE
3/4 A
1/4 A/A

1/4 A/A
1/2 A/a

1/2 A/a
1/4 a/a

1/4 a/a
1/4 a
MENDELIAN ANALYSIS: LONG TREE
2 genes, alleles A, a; B, b... GENOTYPE
1/4 A/A
1/2 A/a
1/4 a/a
1/4 B/B
1/2 B/b
1/4 b/b
1/4 B/B
1/2 B/b
1/4 b/b
1/4 B/B
1/2 B/b
1/4 b/b









1/16 A/A B/B
1/8 A/A B/b
1/16 A/A b/b
1/8 A/a B/B
1/4 A/a B/b
1/8 A/a b/b
1/16 a/a B/B
1/8 a/a B/b
1/16 a/a b/b
MENDELIAN ANALYSIS: LONG TREE
2 genes, alleles A, a; B, b... GENOTYPE PHENOTYPE
1/4 A/A
1/2 A/a
1/4 a/a
1/4 B/B
1/2 B/b
1/4 b/b
1/4 B/B
1/2 B/b
1/4 b/b
1/4 B/B
1/2 B/b
1/4 b/b









1/16 A/A B/B
1/8 A/A B/b
1/16 A/A b/b
1/8 A/a B/B
1/4 A/a B/b
1/8 A/a b/b
1/16 a/a B/B
1/8 a/a B/b
1/16 a/a b/b
9/16 AB
3/16 Ab
3/16 aB
1/16 ab
MENDELIAN ANALYSIS: SHORT TREE
2 genes, alleles A, a; B, b...
PHENOTYPE
¾B

9/16 AB
¼b

3/16 Ab
¾B

3/16 Ab
¼b

1/16 ab
¾A
¼a
much easier... can extend to >2 genes
MENDELIAN ANALYSIS: 1  n GENES
formulae for n genes, dominance, n hybrid crosses …
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
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 a  ½ A+ ½ a
P(a/a) = ½ x ½ = ¼
sum rule: the probability of
A/a x A/a
either of two mutually
½ A+ ½ a  ½ A+ ½ a
exclusive events occurring is
P(A/a) = ¼ + ¼ = ½
the sum of the probabilities of
the individual events... OR
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
YR
Yr
yR
yr

O
315
108
101
32
556
:
9
3
3
1
16
E
313
104
104
35
556
O-E
2
4
-3
-3
(O-E)2
4
16
9
9
2c =
(O-E)2/E
0.013
0.154
0.087
0.257
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
YR
Yr
yR
yr

df
1
2
3
4
5
O
315
108
101
32
556
:
9
3
3
1
16
E
313
104
104
35
556
P .995 .99 .975 .95
.000
.010
.072
.207
.412
.000
.020
.115
.297
.554
.001
.051
.216
.484
.831

.004
.103
.352 >2c>
.711
1.15
O-E
2
4
-3
-3
(O-E)2
4
16
9
9
2c =
(O-E)2/E
0.013
0.154
0.087
0.257
0.511
.9
.75
.5
0.25
.1
.05
.01 .001
.016
.211
.584
1.06
1.61
.102
.575
1.21
1.92
2.68
.455
1.39
2.37
3.36
4.35
1.32
2.77
4.11
5.39
6.61
2.71
4.61
6.25
7.78
9.24
3.84
5.99
7.81
9.49
11.1
6.63
9.21
11.3
13.3
15.1
10.8
13.8
16.3
18.5
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