Mendelian Genetics

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Chapter 3 Mendelian Genetics (Transmission
Genetics)
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Gregor Mendel
1822 born in Heinzendorf 1843 Joined the
Augustinian Monastery of St. Thomas in Brno
(Czech Republic) 1851- 1853 Studied at the
University of Vienna 1854 Returned to Brno and
taught physics and natural sciences 1856 Started
breeding pea plants 1865 Two public lectures on
his findings 1866 Published his work in The
Proceeding of the Natural History
Society 1884 Died of a kidney disorder
His work was largely ignored for 34 years when it
was rediscovered in 1900.
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Mendels experiments are an excellent example of
good scientific technique.

• Chose a well-suited model system (garden peas)
• Carefully designed his experiments
• Collected large amounts of data (very patient!)
• Kept excellent records
• Used mathematical analysis to interpret the
  results (rare in biology at that time)
• The hypothesis was tested by new rounds of
  experiments

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Garden peas are a good model system!

• easy to grow
• cross-fertilized or self-fertilized
• easy to determine source of pollen and egg
• grow fast
• require little space
• produce lots of offspring
• pure-breeding strains for the seven
  characteristics Mendel followed were available

Characteristics (flower color) vs. traits (white,
red)
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Cross-Fertilization of Pea Plants
anther male part stigma ovary female part
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Monohybrid Cross
yellow
green
Pure-breeding
P1
all yellow
all green
X
all yellow
F1
self
F2
6022 yellow 2001 green
Phenotypic ratio
P1 parental generation F1 first filial
generation F2 second filial generation
3 yellow 1 green
(dominant) (recessive)
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Mendels Explanation
P1
GG
gg
X
(yellow)
(green)
G yellow g green
Gg
Gg
X
F1
(all yellow)
self
g
G
F2
GG
Gg
G
gg
Gg
g
Punnett square
Also see Fig. 3.2.
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Experimental Proof of the 121 Genotypic Ratio
Closer look at F2 yellow seeds! Phenotype
yellow genotype GG or Gg
How to tell which one? Self-fertilize the F2
yellow seeds.
X
OR
F2
GG
GG
Gg
Gg
X
all yellow
3 yellow 1 green
F3
1/3
2/3
Thus, if we take the mixture of F2 yellow seeds
and self-fertilize, we should get 1/3 produce
only yellow seeds and 2/3 produce yellow and
green seeds.
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Data from Mendels paper
gg
X
GG
P1
(yellow)
(green)
Gg
F1
(all yellow)
self
8,023 seeds
F2
2001 green (1/4)
6,022 yellow (3/4)
take 519 and self
self
F3
166 yellow only
353 yellow and green
all green
(1 GG)
(2 Gg)
(1 gg)
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Characteristics of Mendels Peas
Fig. 3.1
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Reciprocal Crosses
White
Purple
Purple
White
P1
X
X
100 purple
F1
100 purple
self
self
3 purple 1 white
3 purple 1 white
F2
Note Reciprocal crosses were originally
performed by Josef Gottlieb Kolrulter in the late
1700s.
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Mendels Conclusions
1. Genetic characteristics are determined by
unit factors (genes). 2. Genes exist in pairs
and do not blend. Alternate forms of a gene are
called alleles. 3. Alleles of a gene can be
either dominant or recessive. 4. The two genes
present for each pair of loci randomly segregate
into gametes. (Rule 1 Principle of
Segregation) 5. Fertilization is a random
process.
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Terminology
Gene- A genetic determinant of a trait (e.g. seed
color gene). Allele- Alternate forms of a gene
(e.g. yellow seed or green seed). Can be from 2
to many alleles per gene. Locus- Where a gene is
located on a chromosome. Phenotype- The
observable characteristics of an organism (green
or yellow seeds). Genotype- Genetic composition
of an organism (e.g. GG, Gg, or gg). Homozygote-
An individual having identical alleles in a gene
pair. Heterozygote- An individual having
different alleles in a gene pair.
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Summary of Monohybrid Cross
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Test Cross Monohybrid Cross
Fig. 3.4
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Tips on doing the problems

• Name the genes to relate genotype to phenotype.
  (i.e. G yellow, g green)
• Determine the genotype of the parents.
• Determine the possible gametes.
• Determine the genotypes of the offspring.
• Determine the phenotypes of the offspring from
  the genotypes.

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Sample Problems
You have a plant that has purple flowers and you
know that purple flowers are dominant to white
flowers. How can you tell if this plant is a
homozygote or heterozygote?
W purple w white
Cross with a white plant (must be homozygous
recessive).
Ww x ww
WW x ww
1/2 Ww 1/2 ww
All Ww
purple white
purple
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Sample Problem
Two black guinea pigs were mated and over several
years produced 29 black and 9 white
offspring. How is coat color inherited? Parents
genotype? Offspring genotype?
Name the genes. B black b white
Genotype of parents Bb and Bb
Possible gametes B or b
Genotype of offspring BB Bb Bb bb (121)
phenotype of offspring 3 black 1 white (31)
Coat color is an autosomal trait with black being
dominant to white.
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Dihybrid Cross
phenotypic ratio 9331
Fig. 3.5
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Dihybrid Cross
Genotype ratio 121242121 Phenotype
ratio 9331
G yellow seeds g green seeds W round
seeds w wrinkled seeds
Fig. 3.7
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Rule 2 Independent Assortment
Fig. 3.7
Look at seed color only yellow seeds green
seeds
9/16 3/16 12/16 3/4
3/16 1/16 4/16 1/4
Look at seed coat only round seeds wrinkled
seeds
9/16 3/16 12/16 3/4
3/16 1/16 4/16 1/4
Independent assortment can also account for large
amounts of variation.
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Meiosis accounts for Segregation and Independent
Assortment
Walter Sutton Theodor Boveri
Fig. 3.11
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Test Cross Dihybrid Cross
4th possibility GGWW
Fig. 3.8
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Sample Problems
In dogs, dark coat color is dominant over albino
and short hair is dominant over long hair.
Determine the genotypes of the parents in the
following crosses.
parental pheontypes offspring
phenotypes D,S D,L A,S A,L 1.
dark, short x dark, short 89 31 29
11 2. dark, short x dark, long 18 19
0 0 3. dark, short x dark, short
46 16 0 0 4. dark, short x dark,
long 30 31 9 11 5.
albino, short x albino, short 0 0
28 9
Cross
define the genotypes D dark d albino S
short s long
1 DdSs x DdSs
2 DDSs x DDss
3 DDSs x DDSs
4 DdSs x Ddss
5 ddSs x ddSs
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Rules of Probability
Aa x Aa Probability of a normal offspring being a
carrier?
2/3
Answer
product rule probability of independent events
all happening is the product of their individual
probabilities (and). sum rule When either of
two events can happen the probability can be
calculated by adding the individual probabilities
(or).
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Trihybrid Cross Defining of gametes
gametes 2n
n characteristics
Also see Table 3.1
Fig. 3.9
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Three ways to predict offspring!
1. Punnett Square- Straight forward but
cumbersome when 3 or more characteristics are
looked at.
AaBbCc x AaBbCc
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2. Branch Diagram- Uses product rule. Good for
looking at trihybrid crosses.
phenotypic ratios
monohybrid 31 dihybrid (31)(31)
9331 trihybrid (31)(31)(31)
279993331
Fig. 3.10
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3. Straight Product Rule- Same as branch diagram
except do not write the whole thing out.
AaBbCcDdEE x aaBbCcDdee
What is the probability of the resulting
offspring being aabbCCDdEe ?
(1/2)(1/4)(1/4)(1/2)(1) 1/64
What is the probability of the first parent
producing the gamete abcdE? Answer
(1/2)(1/2)(1/2)(1/2)(1) 1/16
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Pedigree Analysis
Proposius (proband)- Individual who drew your
attention to the pedigree.
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Albinism An autosomal recessive trait
I
Aa
Aa
1
2
heterozygous
II
aa
A-
A-
Aa
AA
AA
4
6
3
5
1
2
III
Aa
Aa
Aa
Aa
A-
A-
5
6
1
2
3
4
homozygous recessive
IV
aa
A-
A-
2
1
3
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Characteristics of an Autosomal Recessive Trait

• males females
• Affected individuals usually have unaffected
  children.
• Most affected individuals have unaffected
  parents.
• Parents of affected individuals are usually
  related.
• An individual with an affected sibling has a 25
  chance of also being affected.
• The trait may skip generations (masked in the
  heterozygous form).

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Huntington disease an autosomal dominant disease
Hh
hh
I
1
2
II
hh
hh
hh
hh
hh
Hh
Hh
4
2
6
3
5
1
7
hh
hh
hh
hh
Hh
Hh
hh
hh
Hh
III
6
5
1
2
3
4
7
8
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Characteristics of an Autosomal Dominant Trait

• males females
• Affected offspring has one affected parent
  (equally likely to be mom or dad, except for new
  mutations).
• Individual with affected parent has 50 chance of
  also being affected.
• Does not skip generations.

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Sample Pedigree Problems
Can two brown-eyed parents have a blue-eyed child?
I
bb
B-
B-
bb
blue eyes brown eyes
3
4
1
2
II
Bb
Bb
Bb
Bb
Bb
Bb
4
5
1
2
3
6
III
bb
B-
B-
1
2
3
Answer YES!!!
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Sample Pedigree Problems (continued)
I
bb
B-
B-
bb
blue eyes brown eyes unknown phenotype
4
2
3
1
II
Bb
Bb
Bb
Bb
Bb
Bb
3
6
4
5
1
2
III
B-
B-
B-
B-
bb
1
2
3
4
5
IV
?
?
1
2
What is the probability of IV-1 having blue eyes?
(2/3)(1/2) 1/3
Now, what if I tell you IV-1 has blue eyes? What
is the probability of IV-2 having blue eyes?
1/2