Transcript Prof. Kamakaka`s Lecture 2 Notes

Prior history
To appreciate Mendels work, one must keep in mind the
prevailing theories of inheritance
Preformationism: the idea that gamete contains an
intact organism, was first proposed in the late 1600s.
A preformed human infant or homunculus contained
within a sperm (Male centric view of the world)
Blending inheritance: essences of both sperm and egg
mixed to form offspring intermediate between the
parents. Strains of plants and animals generated by
blending
Mendel ignored development, focused solely on transmission of traits
1
Darwin and Heredity
Darwin: Among individuals of any species, there are
differences (variations)
Evolution cannot occur unless there are differences
among individuals
If individuals are identical and remain so generation
after generation, there can be no evolution.
Variation is important
Variation must be transmitted from parent to
offspring
Mechanism of inheritance is important for
understanding evolution (1860s)
Mechanism of inheritance not understood
Certain traits suddenly appear in an individual
These traits are then transmitted to their progeny
Porcupine man- protuberances on body
Not due to the environment- only the man and his
offspring have these traits
2
Pangenesis
Pangenesis: (1860s)
Whole organism reproduces itself
Gemmules determine characteristics (traits) of
organism
Germ cells contain gemmules of all sorts and these are
transmitted to the next generation.
Fertilization- gemmules unite and produce new cells of
the types from which they are produced.
3
Cytology
Cytology:
1600s Organisms are composed of cells
1830s The most distinct structure in a cell is the
nucleus. ALL cells have a nucleus. Role of nucleus was
controversial because it disappears during cell
division.
1840s Cells are formed by the division of preexisting cells (Mitosis)
1850s Sperm and Ovum are cells.
1873 Mitosis described in detail- nuclear and
chromosome dynamics described
1883 Fertilization in sea urchin showed that sperm
and ovum fuse. This links parents to offspring
Problem of ploidy:. Nuclei of parents and progeny are
diploids. Are nuclei of germ cells haploid?
1885 Meiosis described- Reductional division of
chromosomes keeps number of chromosomes constant.
No similar phenomenon seen in any other cellular
organelle.
4
Cytology’s contribution to Mendel
Heredity is a consequence of genetic continuity
Germ cells are the vehicle of transmission of traits
from one generation to the next.
Germ cells contain half the number of chromosomes
found in body cells.
Fertilization involves union of sperm and eggs.
Fertilization involves union of nuclei.
Nucleus is crucial. During division it resolves into long
chromosomes that split lengthwise.
Chromosomes do not lose their individuality. They are
inherited intact.
Diploid embryo descends from maternal/paternal fusion
of haploid gametes/chromosomes
5
The origin of genetics:
The study of genetics begins when Gregor Mendel, in
1865, addressed the question :
"How are characters passed on from one generation to
the next?”
Mendel was the first to make a serious attempt of
experimentally answering the question of heredity and
not only were his answers correct, they were a complete
and compelling proof.
Mendel published in 1866 but little attention was paid to
his work until 1900, when it was simultaneously
rediscovered by three scientists, one in Holland, one in
Austria, and one in Germany.
There are often impressions that Mendel was removed
from the scientific community, or that his papers were
not well circulated. This was not true. Over 200 copies
of Mendels papers have been discovered in different
libraries.
All three of Mendel's rediscovers had read Mendel's
work prior to publishing their own work.
6
Gregor Mendel was born on either 20th or 22nd July,
1822 in Heizendorf (today Hynice in the Czech
Republic).
From 1851 to 1853, Gregor Mendel studied zoology,
botany, chemistry, and physics at the University of
Vienna.
He studied botany under Prof. Unger where he learned
genetic crosses
He studied physics under Prof. Doppler where he learnt
statistics
Mendel returned to Brno and began his experiments with
the hybrid cultivation of pea plants in 1856.
After spending eight years carrying out experimental
work in the monastery garden, he reported on the
results of his observations at the meetings of the
Association for Natural Research in Brno on the
evenings of February 8th and March 8th, 1865
Why Peas?
7
Model organisms
A model organism is a species that has been widely studied, usually
because it is easy to maintain and breed in a lab and/or has particular
experimental advantages and are used to obtain information about species
that are more difficult to study directly.
Remember: processes are conserved!!!!!
Genetic model organisms
Amenable to genetic analysis
Breed in large numbers
Short generation time (large-scale crosses over several generations.)
Many different mutants available
Detailed genetic maps
Baker's yeast (Saccharomyces cerevisiae), the fruit fly (Drosophila
melanogaster) and the nematode (Caenorhabditis elegans)
Experimental model organisms
Produce robust embryos that can be studied and manipulated with
ease. Used in developmental biology and biochemistry
Chicken, Zebrafish and Xenopus laevis
Genomic model organisms
Occupy a pivotal position in the evolutionary tree
Special quality of their genome.
An example is the puffer fish (Fugu rubripes) which has a similar
gene repertoire to humans but a much smaller genome (400 million
base pairs instead of 3000 million
Modern genetics:
Location in evolutionary tree
Genome size
Ease of cloning a gene
Ability to insert DNA back into genome
Ability to monitor development
Xenopus, dog, tetrahymena
8
The Pea
Mendel chose the common garden pea to study
patterns of inheritance. This was a excellent choice as
a model system for the following reasons:
1
Many varieties
2
It can self pollinate and cross pollinate
3
Cheap
4
Short generation time
He identified over 20 traits and studied 7
(why 7?)
Seed shape: round versus wrinkled
Seed color: yellow versus green
Flower color: red versus white
Pod shape: inflated versus pinched
Pod color: yellow versus green
Flower position: axial versus terminal
Stem length: long versus short
9
True breeding
The first two years of Mendel's work were devoted to
selecting lines that breed true (pure lines) for a
particular character or trait.
He identified over 20 traits that bred true and
studied 7
Breeding True:
smooth seed
|
Self cross
|
smooth seed
------------------wrinkled seed
|
Self cross
|
wrinkled seed
He identified plants that produced only smooth seeds
and plants that produced only wrinkled seeds.
He identified plants that produced only purple flowers
and plants that produced only white flowers.
10
Mendel’s first cross
P
purple X
male anther
white ----> All purple
female stigma
F1
purple
purple
X
-----> purple:white
705 purple:224 white
------------Reciprocal cross
P
purple
X
male
white
----> purple
F1
purple
x purple -------> Purple:white
female
Blending may not be correct
Reciprocal cross also gave the same result
Therefore preformationism also not correct
11
Mendel’s first cross
P
yellow X
green
----> All yellow
Mendel crossed pure breeding yellow pea plants to pure
breeding green pea plants. All of the progeny were
yellow pea plants. Next he selfed these yellow plants
allowing the pollen to fall on its own stigma.
F1
yellow X
yellow=6022
yellow
----> yellow:green
green=2001
12
Keys to success
He did these experiments with all seven traits!!!!
The ratios obtained were between 2.82:1 and 3.15:1
This cross involving only one character, seed color, is
called a monohybrid cross.
Keys to success:
1
Generated a pure breeding line before starting
the cross
2
Analyzed individual character
3
Counted progeny
13
Conclusions
Although others were doing crosses at the time,
Mendel work was unique
Results and Conclusions:
1
In F1 all progeny showed only one of the two
possible traits for all crosses
2
Males and females not important
3
Character missing in F1, reappeared in F2
Traits did not blend in the offspring but were
transmitted in a discrete fashion and remained
unchanged.
Reciprocal crosses produced the same results, this
indicated that each parent makes an equal contribution
to genetic makeup of the offspring. The sperm does
not contain a homunculus.
14
Terms:
Dominant and recessive: All F1 seeds were yellow but
when selfed the F2 produced some green seeds.
Mendel termed the trait that is expressed in the F1 as
dominant
The trait that is hidden but re-expressed in the F2 as
recessive.
The F1 plants must contain factors for green and yellow
since both reappear in the F2.
------------------------------------------With these assumptions the simplest model is that the
F1 contains two hereditary factors
One for green and another for yellow
We will use the Uppercase Y to represent the dominant
yellow factor and the lower case y to represent the
recessive green.
15
Genotype
The parental yellow pea plants are a pure line- they
only produce yellow pea plants when selfed.
However the F1 yellow pea plants produce some green
pea plants when selfed.
The Yellow of the P generation is different from the
yellow from the F1 generation
If there are Two factors regulating this phenotype- we
can represent these phenotypes as:P
YY
x
yy
(yellow)
(green)
F1
Yy
(yellow)
Yy (yellow) X Yy (yellow)
F2
1YY:
1 Yellow:
2Yy:
2Yellow:
1yy
1Green
It is necessary to make the distinction between the
appearance of an organism and its genetic make-up.
Phenotype refers to the appearance of an organism
Genotype refers its genetic makeup
16
The parental yellow pea plants and F1 yellow pea plants have
the same phenotype but a different genotype.
The Principle of Segregation:
If there are two factors for each trait in an individual, then
without a mechanism to halve the number of factors in each
generation, the factors would multiple with each generation and
become unmanageable.
Mendel therefore reasoned that during gamete formation the
paired factors separate and each gamete receives only one of
the two factors.
Parent
YY
Gametes
Y
F1
yy
Y
y
y
Y y
Sperm and egg then randomly combine to produce F2 progeny
reconstituting two copies of each factor in the progeny
Notice that while the parents have two factors, they produce
gametes containing only a single factor and the progeny again
have two factors
He developed this idea without knowing about meiosis!
17
yy
Green
YY
Yellow
Grows into
A plant
Grows into
A plant
Generates
Gametes
Y
Or
Y
Generates
Gametes
Gametes combine at Random
y
Or
y
Yy
Yellow
Yy
Yellow
Grows into
A plant
Generates
Gametes
Y
Or
y
18
Mendel's assumption of two factors and segregation makes
a strong prediction concerning the genetic make-up of the of
F2 yellow pea plants
F1
Yy
Y
X
y
Y
y
Yy
Y
y
Y
y
YY
Yy
F2
yy
yY
Mendel's prediction:
Phenotype ratio: 3:1 (yellow:green)
Genotype ratio: 1:2:1
1/4YY
Yellow
1/2Yy
yellow
3/4
1/4yy
green
1/4
genotype
Of the yellow F2 plants, 1/3 should be YY and 2/3 Yy
19
How would you test this prediction?
What about the green plants? All should be yy
Selfing
Mendel selfed each of the green F2 plants
Green
yy
F3
X
X
Green
yy
yy
green
120 green F2 were selfed: all 120 plants gave rise ONLY to
green progeny
20
Selfing
Mendel selfed each of the yellow F2 plants
There are two types of yellow YY OR Yy
Yellow
Yellow
Yellow
Yellow
YY
X
YY
Yy
X
Yy
Gametes
Y
Y
Gametes
Y y
Y y
F3
YY
yellow
YY:Yy:yy
3 yellow:1green
1/3 of F2 yellow plants were YY 2/3 of F2 yellow plants were Yy
Of the yellow F2 plants, 1/3 should be YY and 2/3 Yy
That is what Mendel observed!
519 yellow F2 were selfed: 166 gave all yellow and 363 gave
yellow/green
F2 Phenotype ratio: 3:1 yellow:green
F2 Genotype ratio: 1:2:1
1/4YY
Yellow
¾
1/2Yy
yellow
1/4yy
green
¼
genotype
Is there another way to test this prediction?
21
Test cross
If instead of selfing the F2 plants, they are crossed to
pure breeding green plants, what are the expected
outcomes:
F2
YY
x
yy
Yy
x
yy
yy
x
yy
------>all yellow (Yy)
------>1:1 yellow (Yy):green (yy)
------>all green (yy)
22
Genotype Probability
F1
Yy
X
y Y
Y
F
2
Yy
y
Y
y
Y
Y
Y
Y
y
y
y
Y
y
y
Lets cross a Yy with a Yy pea plant
What is the probability of obtaining a homozygous YY plant
Chance of a Y sperm uniting with a Y egg
Chance of sperm with Y allele
1/2
Chance of egg with Y allele
Chance of Y and Y uniting
--------------
1/2
½x½=¼
What is the probability of obtaining a heterozygous Yy plant
Chance of sperm with Y allele and egg with y allele
=(1/2x1/2)=1/4
OR
Chance of sperm with y allele and egg with Y allele=
(1/2x1/2)=1/4
Chance of Yy = 1/4+1/4= 1/2
23
More terms:
Mendel's factors are now known as genes
Alternative forms of a gene that determine A traits are
known as---_alleles__
Individuals with two identical alleles are said to be -----_homozygous_
Individuals with two different forms of alleles are said to
be—
_heterozygous__
24
The dihybrid cross and the principle of independent assortment:
In the second set of experiments Mendel investigated the
pattern of inheritance for two sets of characters
simultaneously.
A cross involving two sets of characters is called a
dihybrid cross.
Pea shape: smooth, wrinkled (Smooth is dominant to
wrinkled)
Cotyledons color: yellow, green (Yellow is dominant to
green)
P
F1
Smooth yellow
x
wrinkled green
smooth yellow
selfed
F2
315 smooth yellow
101 wrinkled yellow
108 smooth green
32 wrinkled green
9
3
3
1
25
9:3:3:1 ----> (3:1) (3:1)
F2
315 smooth yellow
101 wrinkled yellow
108 smooth green
32 wrinkled green
If we examine seed shape only (smooth, wrinkled) and ignore
cotyledon color (yellow, green), in the F2, we expect to find:
3/4 smooth and 1/4 wrinkled:
Take all the smooth peas from the four classes and add them up
# Smooth = 315 + 108 = 423
# wrinkled = 101 + 32 = 133
423:133 is close to the 3:1 expected
Now, if we only examine cotyledon color and ignore seed shape
we expect 3/4 Yellow to 1/4 green.
# Yellow = 315+101
# green = 108+32
416:140 is also close to the 3:1 expected ratio
The wrinkled and green phenotypes INDIVIDUALLY behave as a
standard recessives in a monohybrid cross.
26
The 9:3:3:1 ratio.
The 9:3:3:1 ratio appears a lot more complex than the
3:1 ratios of the monohybrid cross.
Correns insight was to realize the 9:3:3:1 ratio is
nothing more than two 3:1 ratios combined at random.
(3:1) x (3:1)
That is if one examined the traits individually they
formed a 3:1 ratio.
To determine the mode of inheritance of the two genes
in this dihybrid cross Mendel examined each of the
traits separately:
27
Each trait behaves as a standard recessive found in a
monohybrid cross.
They do not affect one another
Genes segregate independently!!!!
GeneS in a gamete does not affect the segregation of
geneY in that gamete
Parent
Gamete
ss yy
SS YY
sy
SY
Ss Yy
F1
Self cross
Ss Yy
x
Ss Yy
28
In a heterozygous individual (Self cross)
F1 self cross
SsYy
x
SsYy
The two genes combine in two ways during gamete formation
This occurs at random giving rise to four classes of gametes
SY
Sy
S
Y
s
y
or
S
y
s
Y
sy
Gamete
sY
SY
sy
Sy
sY
25%
25%
25%
25%
29
Independent assortment
If independent assortment is occurring, four different
kinds of gametes will be produced in equal frequencies.
The only rule is that S and s segregate to separate
gametes and
Y and y segregate to separate gametes
(that is one does not get an Ss gamete or a Yy gamete)
In a group of individuals
SsYy males and SsYy females can produce four types of
gametes in equal frequencies:
SY, Sy, sY, sy
These male and female gametes randomly fertilize each
other to restore diploidy
30
Independent assortment of gene pairs
YYSS x
yyss
YySs
x
YySs
9Y-S-:3Y-ss:3yyS-:1yyss
Different gene pairs assort independently during gamete
formation.
The presence of a Y in a gamete does not influence the
probability of a S or s being in that gamete.
31
Punnet diagram of a dihybrid cross
male
Female
1/4
SY
1/4
Sy
1/4
sY
1/4
sy
1/4
SY
1/4
Sy
1/4
sY
1/4
sy
1/16
SY/SY
Smooth
yellow
1/16
SY/Sy
Smooth
yellow
1/16
SY/sY
Smooth
yellow
1/16
SY/sy
Smooth
yellow
1/16
Sy/sY
Smooth
yellow
1/16
Sy/sy
Smooth
green
1/16
Sy/SY
Smooth
yellow
1/16
sY/SY
Smooth
yellow
1/16
sy/SY
Smooth
yellow
1/16
Sy/Sy
Smooth
green
1/16
sY/Sy
smooth
yellow
1/16
sy/Sy
Smooth
green
1/16
sY/sY
wrinkled
yellow
1/16
sy/sY
wrinkled
yellow
1/16
sy/sY
wrinkled
yellow
1/16
sy/sy
wrinkled
green
9 smooth yellow: 3 smooth green: 3 wrinkled yellow: 1 wrinkled green
Therefore the 9:3:3:1 ratio is a natural outcome of applying
Mendel's two laws
32
Punnet diagram of a dihybrid cross
1/4SY
1/4 SY
1/4 Sy
1/4 sY
1/4 sy
1/16
SY/SY
Smooth
yellow
1/16
Sy/SY
Smooth
yellow
1/16
sY/SY
Smooth
yellow
1/16
sy/SY
Smooth
yellow
1/4Sy
1/16
SY/Sy
Smooth
yellow
1/16
Sy/Sy
Smooth
green
1/16
sY/Sy
smooth
yellow
1/16
sy/Sy
Smooth
green
1/4sY
1/16
SY/sY
Smooth
yellow
1/16
Sy/sY
Smooth
yellow
1/16
sY/sY
wrinkled
yellow
1/16
sy/sY
wrinkled
yellow
1/4sy
1/16
SY/sy
Smooth
yellow
1/16
Sy/sy
Smooth
green
1/16
sy/sY
wrinkled
yellow
1/16
sy/sy
wrinkled
green
33
Significance of Ratios
What is the biological significance of the 9:3:3:1 ratio? This
ratio is only produced if TWO DIFFERENT GENE PAIRS assort
independently of each other during gamete formation.
The presence of one gene in a gamete does not influence
the probability of another gene being found in that gamete
Principle of segregation: for one gene, each individual has two
copies. These two copies segregate from one another during
gamete formation.
Independent assortment: Segregation of one gene pair is
independent of the segregation of any other pair of gene.
34
xxxxxx
35
Mendels laws
Mendel inferred these gamete ratios by selfing
SsYy individuals and looking at phenotypes of progeny.
He could also have inferred these gamete ratios by crossing
SsYy individuals to ssyy individuals.
Crossing to the homozygous recessive individuals is known as a
test cross
36
A test cross is easier that a self cross for the F2
SsYy
x
Gamete:
1/4 SY
1/4 Sy
1/4 sY
1/4 sy
Test cross
ssyy
1 sy
What are the expected genotypic and phenotypic ratios of
the progeny produced from this cross?
1
sy
1/4
SY
1/4
sy
1/4
SY/sy
Smooth
Yellow
1/4
sy/sy
Wrinkled
Green
1/4
sY
1/4
sY/sy
Wrinkled
Yellow
1/4
Sy
1/4
Sy/sy
Smooth
green
What are the expected
genotypic and phenotypic
ratios of the
progeny produced from this
cross?
To answer this, first describe
the gamete classes and their
frequencies produced from
the SsYy and ssyy individuals.
Use these to construct a
Punnett square or branched
diagram
37
38
Rules of Probability
Independent events: The probability of two events
occurring together
What is the probability that both X and Y will
occur together?
Answer: First determine the probability of each event
Then multiply them together.
Mutually exclusive events: The probability of one or another
event occurring.
What is the probability of X or Y occurring?
Answer: First determine the probability of each event
Then add them together.
Probability and Mendel’s Results
Cross Yy xYy pea plants.
What is the chance of Y sperm uniting with a Y egg
½ chance of sperm with Y allele
½ chance of egg with Y allele
Therefore chance of Y and Y uniting = ½ x ½ = 1/4
----------------------------------------What is the chance of Yy offspring
½ chance of sperm with y allele and egg with Y allele
or
½ chance of sperm with Y allele and egg with y allele
Therefore chance of Yy = (½ x ½) + (½ x ½) = 2/4, or
1/2
Yellow is dominant to green
Smooth is dominant to wrinkled
Calculating Phenotype probability
A heterozygous smooth yellow pea is crossed to a
heterozygous smooth yellow pea. (SsYy x SsYy)
What is the expected ratio for a smooth green pea?
1/4SY
1/4 SY
1/4 Sy
1/4 sY
1/4 sy
1/16
SY/SY
Smooth
yellow
1/16
Sy/SY
Smooth
yellow
1/16
sY/SY
Smooth
yellow
1/16
sy/SY
Smooth
yellow
3/4 Yellow
1/4Sy
1/16
SY/Sy
Smooth
yellow
1/16
Sy/Sy
Smooth
green
1/16
sY/Sy
smooth
yellow
1/16
sy/Sy
Smooth
green
1/4sY
1/16
SY/sY
Smooth
yellow
1/16
Sy/sY
Smooth
yellow
1/16
sY/sY
wrinkled
yellow
1/16
sy/sY
wrinkled
yellow
1/4sy
1/16
SY/sy
Smooth
yellow
1/16
Sy/sy
Smooth
green
1/16
sy/sY
wrinkled
yellow
1/16
sy/sy
wrinkled
green
3/4x3/4= 9/16
3/4 Smooth
1/4 Green
3/4 Yellow
1/4 Wrinkled
3/4x1/4= 3/16
1/4x3/4= 3/16
1/4 Green
1/4x1/4= 1/16
41
Probability
Yellow is dominant to green
Smooth is dominant to wrinkled
Heterozygous yellow
Heterozygous smooth
X
Heterozygous yellow
Homozygous wrinkled
Probability of yellow wrinkled?
1/2 smooth
3/4 yellow
1/2 wrinkled
1/4 green
Heterozygous yellow
Homozygous smooth
X
Heterozygous yellow
Homozygous wrinkled
Probability of yellow wrinkled?
1 smooth
3/4 yellow
0 wrinkled
42
1/4 green
Calculation
Red is dominant to white eye
Flat wing is dominant to delta wing
A heterozygous Red eyed flat winged fly is crossed
to a heterozygous Red eyed flat winged fly.
What is the expected ratio for a Red eyed flat
winged fly?
43
xxxxxx
44
More Phenotype probabilities
Yellow is dominant to green
Smooth is dominant to wrinkled
Tall is dominant to short
Heterozygous yellow
Heterozygous smooth
Heterozygous tall
X
Heterozygous yellow
Homozygous wrinkled
Heterozygous tall
Probability of green wrinkled short?
3/4 tall
1/2 smooth
1/4 green
1/2 wrinkled
1/4 short
3/4 tall
1/4 short
3/4 tall
1/2 smooth
3/4 yellow
1/4 short
3/4 tall
1/2 wrinkled
45
1/4 short
Genotype probability
P
YYRRTTSS  yyrrttss
F1
YyRrTtSs  YyRrTtSs
What is the probability of obtaining the genotype
YyRrTtss in the F2?
-Four loci
Each loci is heterozygous i.e. 2 types of gametes
- 24 = 16 possible gamete combinations for each parent
Thus, a 16  16 Punnet Square giving rise to 256
genotypes
F1
gametes
YyRrTtSs
X YyRrTtSs
RYTS RYTs RYtS RYts
RYTS RYTs RYtS RYts
RyTS RyTs RytS Ryts
RyTS RyTs RytS Ryts
rYTS rYTs rYts rYTS
rYTS rYTs rYts rYTS
ryTs rYtS
ryTs rYtS
rYts
ryts
rYts
What is the ratio of different genotypes in the F2?
ryts
Branched diagram for genotypes
F1
Alternatively –
Loci Assort Independently
Look at each gene locus separately.
X
Yy
y Y
Y
F
2
P
Yy
y
Y
y
Y
Y
Y
Y
y
y
y
Y
y
y
YYRRTTSS  yyrrttss
F1
YyRrTtSs  YyRrTtSs
What is the probability of obtaining the genotype
YyRrTtss in the F2?
Yy X Yy
Rr  Rr
1YY:2Yy:1yy
1RR:2Rr:1rr
2/4 Yy
2/4 Rr
Tt  Tt
Ss  Ss
1TT:2Tt:1tt
1SS:2Ss:1ss
2/4 Tt
1/4 ss
Probability of obtaining individual with Rr and Yy and Tt and ss.
2/4  2/4  2/4  1/4 = 8/256 (or 1/32)
Genotype probability
P
RRYYTTSS  rryyttss
F1
RrYyTtSs  RrYyTtSs
What is the probability of obtaining a completely homozygous
Dominant or completely homozygous recessive genotype in F2?
Genotype could be RRYYTTSS or rryyttss
Rr  Rr
Yy  Yy
Tt  Tt
Ss  Ss
1RR:2Rr:1rr 1YY:2Yy:1yy
1TT:2Tt:1tt
1SS:2Ss:1ss
1/4 RR
1/4 YY
1/4 TT
1/4 SS
1/4 rr
1/4 yy
1/4 tt
1/4 ss
(RR x YY x SS x TT) + (rr x yy x ss x tt)
(1/4  1/4  1/4  1/4) + (1/4  1/4  1/4  1/4)
= 2/256
Specific breeds of dogs are also associated with specific diseases
Dobermans: narcolepsy
Scotties: Haemophilia
Terriers: copper metabolism (menke disease)
Labrador: hip dysplasia
Beagle: seizure risk
The ratio from a doberman cross suggests that narcolepsy is most
likely mediated by a single gene! (3:1 ratio)
Why are Mutts healthier than true breeds?
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Hybrid Vigor
Hybrid vigor:
The first cross between two purebred lines is often healthier than
either parent
Breed1
a-B-C-d-E
a-B-C-d-E
F1
Breed2
A-b-c-d-E
A-b-c-d-E
a-B-C-d-E
A-b-c-d-E
If GeneA causes narcolepsy and geneC causes haemophilia
F1 will be normal.
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Corn
Commercial corn is a F1 hybrid because of hybrid vigor
Two inbred lines are mated to generate a F1 that is sold
AAbb
F1
x
aaBB
AaBb
hybrid vigor
1930’s increase in corn output because of these hybrid varieties
Line1 -disease resistant- rust and mold resistance but low yield
Line2- large crop but susceptible to disease
Hybrid- disease resistant and large yield----Sold to farmer
The F1 hybrids do not breed true!
The farmer cannot plant the seeds he gets from the F1 hybrid in the
next season. They will not be disease resistant with high yields!
Gametes AB
Ab
aB
ab
The seeds will generate plants with
A-Bhigh yield, disease resistant
A-bb
high yield, susceptible to disease
aaBlow yield, disease resistant
aabb
low yield, susceptible to disease
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