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Ch3 Mendelian Genetics
You are responsible
for reviewing the
content of this
chapter.
Chapter CONCEPTS
• Inherited characteristics are under the control
of discrete particulate factors called genes that
are transmitted from generation to generation
on vehicles called chromosomes, according to
rules, first described by Gregor Mendel The
outcomes of crosses subject to these rules are
affected by chance deviation and can be
evaluated using statistical analysis. Human
traits are initially studied using the pedigree
method .
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3.1 Gregor Mendel
3.2 Monohybrid cross
3.3 Dihybrid cross
3.4 Trihybrid cross
3.5 Rediscovery of Mendel’s Work
3.6 Independent Assortment and genetic Variation
3.7 Probability and Genetic Event
3.8 Evaluating genetic data :Chi-square
3.9 Human Pedigree
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 pollination:
self and cross fertilization
1. Gregor Mendel
• Genotype and Phenotype (study figure 3.1)
• Mendel’s Experimental Design
– Understand the difference between selffertilization and cross-fertilization.
– Understand the organism used in Mendel’s
design.
– Understand what true-breeding implies.
–
testcross
– Understand the 7 characteristics or traits that
Mendel examined & alternative forms of each.
3.2 Monohybrid cross
• Monohybrid Crosses & Law of segregation
– Understand what the notations for the different
genotypes are.
– Understand what monohybrid crosses are and what
reciprocal crosses are.
– Understand all the terminology in this section
Understand what the Law of Segregation means
and what genotypic & phenotypic ratios are.
– Natural segregation
3.3 Dihybrid cross
• Dihybrid Crosses & Law of Independent
Assortment
– Understand what the Law of Independent Assortment
means and predicts in terms of dihybrid phenotypic &
genotypic ratios.
– Understand how to use the branch diagram for
dihybrid crosses & how to do dihybrid testcrosses.
– Understand how to predict results of trihybrid or
greater crosses using the branch diagram.
How Mendel's Peas Become Wrinkled:
A Molecular Explanation
• Only recently, well over a hundred years after
Mendel used wrinkled peas in his ground-breaking
hybridization experiments, have we come to find out
how the wrinkled 皱缩gene makes peas wrinkled.
• The wild-type allele of the gene encodes a protein
called starch-branching enzyme (SBEI). This
enzyme catalyzes the formation of highly branched
starch molecules as the seed matures. Wrinkled peas,
which result from the homozygeus presence of the
mutant form of the gene, lack the activity of this
enzyme.
• The production of branch points is inhibited during
the synthesis of starch within the seed.
• This in turn leads to the accumulation of more sucrose
and to a higher water content while the seed develops.
• Osmotic pressure inside rises, which causes the loss of
water internally and ultimately the wrinkled
appearance of the seed during its maturation.
• In contrast, developing seeds that bear at least one
copy of the normal gene (being either homozygous or
heterozygous for the dominant allele) synthesize
starch, and reach an osmotic balance that minimizes
the loss of water
• the end result is a smooth-textured outer coat
• The SBEI gene has been cloned and analyzed.
Interestingly, the mutant gene contains a foreign,
sequence of some 800 base pairs that disrupts the
normal coding sequence.
• This foreign segment closely resembles other such
sequences, called Transposable elements .
• They have the ability to move from place to place in
the genome of organisms.
• Transposable elements similar to this one have been
found in maize (corn), parsley, and snapdragons.
• Study of the SBEi gene provides greater insight into
the relationship between genotypes and phenotypes.
• We cover transposable elements greater detail in
Chapter 14.
3.4 Trihybrid cross
• Three factor cross
– Understand how to use Punnett squares and the
branch diagrams.
– Understand what probability is and understand
when to use the product rule and sum rule of
probability.
– Understand what a testcross is and the expected
Mendelian phenotypic & genotypic ratios
produced in testcrosses.
MENDELIAN ANALYSIS:
1  n GENES
3 methods of working out expected outcomes of
controlled breeding experiments:
1. Punnet square
2. tree method (long) – genotypes
. tree method (short) – phenotypes
3. (a+b)N
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
¼ A/A
 1/4 A/A
½ A/a
 1/2 A/a
¼ a/a
 1/4 a/a
PHENOTYPE
3/4 A
1/4 a
MENDELIAN ANALYSIS: LONG TREE
2 genes, alleles A, a, B, b... GENOTYPE
¼ A/A
½ A/a
¼ a/a
¼ B/B
½ B/b
¼ b/b
¼ B/B
½ B/b
¼ b/b
¼ B/B
½ B/b
¼ b/b









PHENOTYPE
1/16 A/A B/B
9/16 AB
1/8 A/A B/b
1/16 A/A b/b3/16 Ab
1/8 A/a B/B
1/4 A/a B/b
1/8 A/a b/b
1/16 a/a B/B
3/16 aB
1/8 a/a B/b
1/16 a/a b/b
1/16 ab
MENDELIAN ANALYSIS: SHORT TREE
2 genes, alleles A, a, B, b... PHENOTYPE
¾B
 9/16 AB
¼b
 3/16 Ab
¾B
 3/16aB
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 = ¼ n
4. # possible different phenotypes of progeny = 2n
STATISTICS: BINOMIAL EXPANSION
diploid genetic data suited to analysis (2 alleles/gene)
product and sum rules apply
examples... coin flipping
n
use fomula: (p+q)n =  [n!/(n–k)! k!](p)n–k (q)k =
k=0
define symbols...
From 3.7 Probability and Genetic Event
e.g., outcomes possible from monohybrid cross
1 offspring, n = 1,
(p+q)1 = 1
2 offspring, n = 2,
(p+q)2 = 1
3 offspring, n = 3,
(p+q)3 = 1
(1!/1!0!)(¾)1(¼)0 = 3/4
(1!/0!1!)(¾)0(¼)1 = 1/4
 = 1
(2!/2!0!)(¾)2(¼)0 = (1)9/16
(2!/1!1!)(¾)1(¼)1 = (2)3/16
(2!/0!2!)(¾)0(¼)2 = (1)1/16
 = 1
16/16
(3!/3!0!)(¾)3(¼)0 = (1)27/64
(3!/2!1!)(¾)2(¼)1 = (3)9/64  P of 2A_ + 1aa
(3!/1!2!)(¾)1(¼)2 = (3)3/64
(3!/0!3!)(¾)0(¼)3 = (1)1/16
 = 1
STATISTICS: BINOMIAL EXPANSION
Q: True breeding black and albino cats mate
and have a little of all black kittens. If these
kittens grow up and breed among themselves,
what is the probability that at least two of the
F2 kittens will be albino?
STATISTICS: BINOMIAL EXPANSION
Q: True breeding black and albino cats mate and
have a little of all black kittens. If these kittens grow up
and breed among themselves, what is the probability
that at least two of the 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
P1:
genotype:
gametes:
black
BB
P(B) = 1
×

albino
bb
P(b) = 1
F1:
genotype:
gametes:
black
Bb
P(B) = ½ , P(b) = ½
×

black
Bb
P(B) = ½ , P(b) = ½
expected F 2:
P(black) = P(B_) = (½ )2 + 2(½ )2 = ¾ = p
P(albino) = P(bb) = (½ )2 = ¼ = q
STATISTICS: BINOMIAL EXPANSION
possible outcomes...
F2
*
3 black + 0 albino

P (   ) = (¾ )3 × (¼ )0 = 27/64

P (   ) = (¾ )2 × (¼ )1 = 9/64
P (   ) = (¾ )2 × (¼ )1 = 9/64
P (   ) = (¾ )2 × (¼ )1 = 9/64
1 black + 2 albino

P (   ) = (¾ )1 × (¼ )2 = 3/64
P (   ) = (¾ )1 × (¼ )2 = 3/64
P (   ) = (¾ )1 × (¼ )2 = 3/64
0 black + 3 albino

P (   ) = (¾ )0 + (¼ )3 = 1/64
2 black + 1 albino
Total
Probability
64/64
STATISTICS: PASCAL’S TRIANGLE
Sample Frequency
Possible Outcomes*
Combinations
1×
1+1
2 orders

2×
1+2+1
3 orders
 
3×
4 orders
+++
  
4×
1+4+6+4+1
5 orders
   
5×
1 + 5 + 10 + 10 + 5 + 1

6×



6 orders

1 + 6 + 15 + 20 + 15 + 6 + 1
7 orders
STATISTICS: BINOMIAL EXPANSION
binomial expansion of (p + q)n =
n
 [n!/(n–k)! k!](p)n–k (q)k =
k=0
... if p = ¾ and q = ¼ , ...
expansion of (p + q)3 = (¾ + ¼)(¾ + ¼ )(¾ + ¼ ) =
n
 [n!/(n–k)! k!] (¾ )n–k (¼ )k =
at least two albinos…
k=0
(3!/3!0!) (¾
(3!/2!1!) (¾
(3!/1!2!) (¾
(3!/0!3!) (¾
)3 (¼ )0 =  × 27/64
)2 (¼ )1 =  × 9/64
)1 (¼ )2 =  × 3/64
)0 (¼ )3 =  × 1/16
64/64
0 albinos
1 albino
2 albinos
3 albinos
> 10/64 = 5/32
Binomial Expansion
• 1、遗传分析中后代组合的概率推算
Aa X aa杂交——二个后代出现Aa和aa的组合如
下：
No1.
No2
概率
Aa
Aa
¼
Aa
aa
¼
Aa
Aa
¼
aa
aa
¼
分布方式1：2：1，
正是（p+q)2=1p2+2pq+1q2
[(Aa) + (aa)]2 = 1(Aa)(Aa):2(Aa)(aa):1(aa)(aa)
Aa X aa产生三个后代
• （p+q)3 = (1/2 + ½ )3
基因型及组合
概率
Aa
aa
3
0
P3=(1/2)3=1/8
2
1 3p2q=3(1/2)2(1/2)=3/8
1
2 3pq2=3(1/2)(1/2)2=3/8
0
3
q3=(1/2)3=1/8
一般说，设p为某一基因型（或表型）出现的概率，
q或（1-p)为另一基因型（或表型）出现概率，那
么有 p+q=1。这些事件每一特定组合的概率都可
以用二项式展开说明，二项式展开的系数表示在
总共可能出现的组合数中的比率。
• 杨辉三角：则为二项式展开的系数
1 1
1
1 2 1
2
1 3 3 1
3
1 4 6 4 1
4
1 5 10 10 5 1 5
1 6 15 20 15 6 1 6
s
n-s
n!/s!(n-s)p q
• 如果不考虑出现顺序，可有如下通式计算表
型或基因型某一特定组合的概率：
–
–
–
–
–
n——子代数目
S——某一基因型（或表型）数
n-s——另一基因型（或表型）数
P——某一基因型（或表型）出现的概率
q——另一基因型（或表型）出现的概率
• 例：某医院一天出生6个婴儿，三男三女的概
率是多少？
– （6！/3！3！）（1/2）3（1/2）3=5/16
2、孟德尔遗传的分离比率的二项式
• F2基因型比为（1∶2∶1）2，即（1/4+2/4+1/4）2三
项式展开式的各项系数；
• F2表型比为（3∶1）2 ，即（3/4+1/4）2二项式展开
式的各项系数。
（3：1）n（n—基因对数），既（a+b)2=a2 + 2ab + b2
• 以3代a,以1代b，则有，
– 当两对基因时
(3+1)2=(3)2 + 2(3)(1)+ (1)2
– 当三对基因时
(3+1)3=(3)3 + 3(3)2(1) + 3(3)(1)2 + (1)3
9:9:9
3:3:3
由此可见任何对基因杂种自交后代表型都可以用二项式展开计算。
N对基因的二项式展开
• 随着基因对数的增加，二项式展开将变
得复杂。当有n对基因时：
• (a+b)n=Cn0 an+Cn1 a n-1 b + Cn2a n-2
b2…Cnka n-k bk +Cnnbn
• Cnk = n!/K!(n-k)!
同样可以用杨辉三角推得
3.5 Rediscovery of Mendel’s Work
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
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
3.6 Independent Assortment and
genetic Variation
• One major consequence of independent
assortment is the production by an individual of
genetically dissimilar gametes. Genetic variation
results because the two members of any
homologous pair of chromosomes are rarely if
ever genetically identical. Therefore, because
independent assortment leads to the production of
al1 possible chromosome combinations, extensive
genetic diversity results.
• We have seen that the number of possible gametes, each
with different chromosome compositions, is 2n, where n
equals the haploid number. Thus, if a species has a
haploid number of 4, then 24 or l6 different gamete
combinations can be formed as a result of independent
assortment.
• Although this number is not high, consider the human
Species, where n = 23, if 223 is calcu1ated, we find that
in excess of 8 x l06. or over 8 million, different types of
gametes are represented. Because fertilization represents
an event involving only one of approximately 8 x l06
possible gametes from each of two parents. each
offspring represents only one of (8 x 106)2. or 64 x 1012,
potential genetic combinations!
• No wonder that except for identical twins,
each member of the human species
demonstrates a distinctive appearance and
individuality. This number of combinations is
far greater than the number of humans who
have ever lived on earth !
• Genetic variation resulting from independent
assortment has been extremely important to
the process of evolution in all organisms.
3.7 Probability and Genetic Event
 review essential Mendel
 genetic ratios & rules
 statistics
 chi-square
 binomial expansion
 Poisson distribution
 pedigree analysis
to some front page
3.8 Evaluating genetic data :
Chi-square
• Mendel's 3:1 monohybrid and 9:3:3:1 dihybrid ratios
are hypothetical predictions based on the following
assumptions:
• (1) Each allele is dominant or recessive;
• (2) segregation is operative;
• (3) independent assortment occurs; and
• (4) fertilization is random. The last three assumptions
are influenced by chance events and are therefore
subject to random fluctuation. This concept, called
chance deviation, is most easily illustrated by tossing a
single coin numerous limes and recording the number
of heads and tails observed.
• Chi-square test.
– Understand what a null hypothesis is and
how to state a testable hypothesis.
– Understand what degrees of freedom are.
– Understand how to calculate a chi-square
value and how to interpret what it means.
– Understand how to determine whether to
accept or reject your hypothesis.
– Understand how to read a chi-square table to
interpret the probability of getting a given
chi-square value.
STATISTICS: CHI-SQUARE ANALYSIS
4 phenotypes ~
roughly 9:3:3:1
ratio
hypothesis
what does it
mean ?
test hypothesis
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% (in other words, the
biological explanation is supported by data)
3.9 Human Pedigree
– Understand what a proband is in a pedigree.
– Understand how to recognize dominant and
recessive traits in pedigrees.
– Understand how to predict the probability of an
affected individual appearing in a future
generation.
• Work the problems at the end of the chapter
• Mendelian Genetics in Humans
– Understand the major symbols used in pedigrees.
• Open squares and open circles represent males
and females unaffected by whatever condition is
being studied in the pedigree.
• Shaded squares and shaded circles represent
males and females affected by the condition
studied in pedigree.
• See figure 10.18 (p. 272) for other major symbols
PEDIGREE ANALYSIS
 order of events for solving pedigrees
1. identify all individuals according to 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
• There are 2 major systems used in genetics to
symbolize two alternative forms (alleles) of genes.
– Corn Notation System (used by Mendel)
• Upper case letter is used to symbolize the
dominant allele.
• Lower case form of the same letter is used to
symbolize the recessive allele.
• Since there are more than 26 traits in any
given species, more than one letter may be
used to designate the alleles.
• Example 1: let R = round seeds and r = wrinkled
seeds. Therefore, genotypes are: RR or Rr for
round and rr for wrinkled.
• Example 2: let PU = purple flowers and pu = white
flowers. Therefore, genotypes are: PUPU or PUpu
for purple & pupu for white.
– Drosophila Notation System (used by Morgan)
• Uses same case of a letter(s) for both dominant and
recessive alleles.
• If mutant is recessive to normal, lower case letters
are used for both alleles, and + symbol is added to
the normal (non-mutant) allele.
• If mutant is dominant to normal, upper case
letters are used for both alleles, and + symbol
is added to the normal (non-mutant) allele.
• Example 1: Let b = black body and b+ =
normal body color. Therefore, the genotypes
are b+b+ or b+ b for normal and bb for black
body. Black body is recessive to normal body
color.
• Example 2: Let B = bar eyes and B+ = normal
eyes. Therefore, the genotypes are BB or B B+
for bar eyes and B+B+ for normal eyes. Bar
eyes are dominant to normal eyes.