Transcript Mendelian Genetics - Moodle

Simple Genetics
https://pageslap.wordpress.com/2010/02/
Mendelian Genetics
•Vocabulary
•Mendel’s experiments
•The monohybrid cross
•The dihybrid cross
•Mendel’s laws
•Solving Genetics problems
•Non-Mendelian Genetics
•From genotype to phenotype
Key Concepts
In many species, individuals have two alleles of each gene. The
principle of segregation of alleles states that prior to the
formation of eggs and sperm, the alleles of each gene separate so
that each egg or sperm cell receives only one of them.
(Genes are located on chromosomes…)
The principle of independent assortment of chromosomes states
that alleles of different genes are transmitted to egg cells and
sperm cells independently of each other.
There are important exceptions and extensions to the basic
patterns of inheritance that Mendel discovered.
© 2011 Pearson Education, Inc.
Key Concepts
(…Genes are located on chromosomes. The principle of
segregation is explained by the separation of homologous
chromosomes in anaphase of meiosis I. The principle of
independent assortment applies to genes found on different
chromosomes and is explained by chromosomes lining up
randomly in metaphase of meiosis I.)
© 2011 Pearson Education, Inc.
Some basic terminology
• Chromosome – a long strand of DNA containing
•
•
•
•
•
•
many genes
Gene – a segment of DNA that codes for a protein
Locus – The place on a chromosome where a specific
gene resides – virtually synonymous with gene
Allele – a variant form of a gene (symbolized by, e.g.,
“A” vs “a”, but real difference is in DNA sequence)
Genotype – the combination of alleles an individual
bears at one or more loci
Trait – a general feature that can take on many values
– e.g., eye color
Phenotype – the specific value an individual
expresses for a trait – e.g., blue eyes or brown eyes
Trait
• Hair color
• Eye Color
• Height
• Blood type
vs.
Phenotype
Brown, Red, Black
Brown, Blue, Green
1.0 m, 1.5m, 1.8 m, 2m
A, B, AB, O
Mendelian genetics
• Gregor Mendel
• Mendelian genetics =
discrete, qualitative traits
with simple dominant /
recessive relationship
– Flower color – red or white
– Seed shape – smooth or
wrinkled
Quantitative genetics
• Continuously varying
traits – e.g., height,
milk yield, skin color
• Environmental effects
• Photo: R. A. Fisher
Population genetics
• Changes in allele
frequencies in
populations
• Photo: G. H. Hardy
Genetics before Mendel
• Heredity was obvious
• Blending Inheritance
– Problems with blending
inheritance
• Why not “beige”?
• Loss and recurrence of
phenotypes
• Inheritance of Acquired
Characteristics
• Particulate Inheritance
Why Mendel succeeded
• Selected discrete traits
• Followed several
generations
• Understood probability
• Large sample size
• Photo: Mendel’s garden
Why Mendel succeeded
• Started with pure-breeding parents
(homozygous)
• Carefully controlled matings
The Mendelian hypothesis
• Mathematical model
– Two parents each contribute something
(simplest assumption is one piece of info from
each)
– Each individual must thus have at least two
pieces of info (one from each parent)
– Two pieces can either be the same or
different (same if breed true)
– Which piece an individual passes on is
random, each with a probability of 0.5
Prediction based on probability
• True-breeding parents (P) can each only pass
•
•
on one type of information (homozygous)
Offspring of this first step in the monohybrid
cross (the F1) must thus all have two different
types of information (heterozygous)
Offspring of two F1 (the F2) have 50:50 chance
of receiving one or the other piece of
information from each parent
The product rule
• The probability of two independent random events
co-occurring is the product of their individual
probabilities (probability of A and B)
– If parents are heterozygous (Aa),
• Probability of getting A from mother is 0.5
• Probability of getting A from father is 0.5
• Probability of getting AA is 0.5 x 0.5 = 0.25
• Same for Aa, aA, and aa
The sum rule
• The probability of any of two or more
mutually exclusive events occurring is
the sum of their individual probabilities
(probability of A or B happening)
• Example: If both parents heterozygous,
the probability of receiving either Aa (A
from mom, a from dad) or aA (a from
mom, A from dad) is 0.25 + 0.25 = 0.5
Illustration: Coin tossing
Experiment: The monohybrid cross
• Experiment follows
from predictions
• Cross true-breeding
parents
–Produces uniform F1
• Cross two F1
–produce 3:1 ratio of
phenotypes
–¾ tall, ¼ short
Interpretation
• F1 all heterozygous
• ¼ of F2 Homozygous dominant – TT
• ½ of F2 Heterozygous – ¼ Tt and ¼ tT
– All of above are tall
• ¼ of F2 Homozygous recessive – tt
– These are short
• One allele, when present, masks the effect
of the other
– In this case, T masks the effect of t
Dominant and recessive
• Dominant allele (upper case) masks the
effect of recessive, thus dominant
phenotype is expressed in both
homozygous dominant and heterozygous
genotypes
• Recessive allele (lower case) is masked
by dominant allele, thus recessive
phenotype is expressed only in
homozygous recessive genotypes
• NOTE: this has nothing to do with which
is more common, more beneficial, etc
Genotypic and Phenotypic
ratios in a monohybrid cross
•Genotypic ratios of the progeny of two F1
1TT:2Tt:1tt
or
(0.25TT : 0.5Tt : 0.25tt)
hybrids are:
•If T is dominant to t, phenotypic ratios are:
3 dominant: 1 recessive
or
0.75 dominant : 0.25 recessive
Mendel’s 1st Law (or “Principle”)
segregation of alleles
The two members of each pair of alleles separate or
“segregate” during gamete formation, such that
each gamete contains only one of the pair of alleles
Or
Each of the two alleles an individual carries at a
particular locus will appear in half of that individual's
gametes - and thus be passed on at random to half
of their offspring.
Punnett Squares
• Simplest tool for predicting the outcome of a
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•
•
•
genetic cross is a Punnet Square
Example: F1 Hybrid cross
Aa male X Aa female
Place possible alleles from
one of parents on top; the other
set from the other parent on the
side. Location doesn’t matter.
Fill across from sides and down
from top to find genotypic
combinations in offspring
A
a
A
AA
A a
a
aA
a a
Example: Monohybrid cross
Black female (BB)
P
• Start with true breeding parents
generation
1. What types of gametes can each
produce?
• Black ⇒ only black
• Brown ⇒ only brown
F
2. What are the genotypes in the F1?
generation
• All Bb
B - black
b - brown
3. F1 tells us what?
• Black is dominant to brown
4. What are the phenotypic ratios Gametes
in the F2?
• 3 black:1 brown
5. What are the genotypic ratios in
the F2?
• 1BB:2Bb:1bb
1
Brown male (bb)
Gametes
All Bb
Gametes
F2 generation
Dominant B
masks
recessive b
The test cross
• Used to reveal unknown genotype
–Is a black guinea pig homozygous
or heterozygous?
• Cross dominant phenotype with
recessive phenotype
–Always!
• Alleles obtained from dominant parent
are always expressed in progeny
A testcross is a hypothesis test
If homozygous
Black (BB)
Homozygous
Brown (bb)
If heterzygous
Black (Bb)
Homozygous
Brown (bb)
Gametes
Gametes
All Bb
Eggs
Sperm
All offspring are black and heterozygous
Two unique predictions:
•All Black
•Half black:half brown
Confidence and the test cross
• How many brown offspring would convince
you of the unknown genotype?
• How many black?
– P(2 black if heterozygous)= 0.5 x 0.5 =0.25
– P(3 black if …)= 0.5 x 0.5 x 0.5 = 0.125
• When do you become convinced?...
“Statistically significant”
• Usually defined as p<0.05
• I.e., when an observed event is likely to
happen by chance less than one time in
20 trials if null hypothesis is correct
• In a test cross, P(5 dominant offspring if
heterozygous) = 0.55 = 1/32 = 0.03125
The dihybrid cross
• Having demonstrated pattern of
inheritance for single traits. Mendel asked
how different traits influenced one another
• Experiments similar except started with
parents that bred true for two different
traits
• Question: Does the inheritance of one trait
influence the inheritance of another trait?
– E.g., Does the seed color one inherits
influence the seed shape?
Dihybrid cross continued
• What do you predict if each trait is
independent of other traits?...
• Remember product rule?...
– Probability of two independent events
co-occurring is the product of their
independent probabilities
– Apply this using phenotypic ratios from
two monohybrid crosses, e.g., seed
color and seed shape
Dihybrid cross continued
• If yellow (Y) is dominant to green (y) and
smooth (S) is dominant to wrinkled (s)…
• What is the proportion of yellow offspring
from a mating between two Yy
heterozygotes?
– 0.75 or 3/4
• The proportion of wrinkled from a cross
between two Ss heterozygotes?
– 0.25 or 1/4
Dihybrid cross continued
• So, if P(yellow) = 0.75
and P(wrinkled) = 0.25
and color and shape are independently
inherited…
• What is the proportion of yellow wrinkled
from a cross between two SsYy double
heterozygotes?
= 0.75 x 0.25 = 0.1875 or 3/4 x 1/4 = 3/16
Other combinations?
• Yellow and smooth?
•
•
•
•
– 0.75 x 0.75 = 0.5625 or ¾ x ¾ = 9/16
Green and smooth?
– 0.25 x 0.75 = 0.1875 or ¼ x ¾ = 3/16
(same as yellow wrinkled)
Green and wrinkled?
– 0.25 x 0.25 = 0.0625 0r ¼ x ¼ = 1/16
So if seed color and seed shape are independently
inherited (Note: this is a null hypothesis!), then we
expect a 9:3:3:1 ratio in the offspring of two double
heterozygotes (and this is a prediction based on a
mathematical model)
Can you think of an alternative hypothesis?
Black,
short-haired
A dihybrid cross
P
generation
BBSS
• Parental generation:
•
true-breeding black shorthair
and brown longhair
F1 all Black shorthair
Brown,
long-haired
bbss
Gametes
F1
generation
All BbSs
Gametes formed by segregation and
independent assortment of alleles
– Which tells us what?
• F2 reveals four phenotypes
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–
–
9
3
3
1
black short
black long
brown short
brown long
• Data support Mendel’s
hypothesis!
F2 generation
F2 phenotypes
Mendel’s Law of Independent
Assortment (of Chromosomes)
• Pattern of segregation of alleles for one
trait is independent of the segregation of
alleles for other traits
Or
• Allele received at one locus has no effect
on the allele received at another locus
• This is only true if loci are on different
chromosomes
Can still use a
Punnett square
•List all possible gametes
for the two parents
•This time, you must keep
track of alleles at two loci
•Fill down columns and
across rows to reveal
two-locus diplod
genotypes
•Determine phenotypes
from genotypes
• Punnett squares eventually become
cumbersome (with 3 loci  8x8 =64 cells)
• But product rule and sum rule still apply
• E.g., what is the probability of getting a
SsyyZZ offspring from a cross between
two SsYyZz individuals?
– P(Ss) = 0.5
– P(yy) = 0.25
– P(ZZ) = 0.25
– P(SsyyZZ) = 0.5 x 0.25 x 0.25 =
or
2/4 x 1/4 x 1/4 =
0.03125
2/64
Meiosis explains Mendel’s Laws
• Mendel’s work largely ignored until 1900
• Hugo DeVries, Karl Correns and Erich von
Tschermak independently recognized
significance of Mendel’s work
• W. S. Sutton observed meiosis and made
the connection between it and Mendel’s
laws in 1903
Meiotic basis of
segregation of
alleles
• Segregation occurs
•
during MI when
homologs separate
(or during MII when
chromatids separate if
crossed over)
As a result, half the
ova or sperm contain
the paternal (or
maternal) homolog
and thus the allele it
carries
Meiotic basis of Independent
Assortment of Chromosomes
• The homologs
•
are arranged at
random at the
equatorial plane
in Metaphase I
For 2
chromosomes
with 2 alleles
alleles on each,
this results in
four different
possible
genotypes in the
gametes, in a
1:1:1:1 ratio
Option 1
See Fig 10-8
Option 2
Fig. 11.7
Fig. 11.10
Without crossing over, can generate 2n
different types of gametes
Linkage
• Independent assortment only applies to 2 loci
when they are on different chromosomes.
• However, each chromosome has many loci,
so some pairs of loci will share a chromosome
• In this case, alleles at these loci do not assort
independently of one another
• When 2 loci are on the same chromosome,
we refer to them as being linked to one
another
• Recombinants will be rarer than
parentals
Sex Chromosomes
• The chromosomes that determine individual’s
sex are called the sex chromosomes.
•Other chromosomes are
called the autosomes.
Humans have 22
autosomes and females
have a pair of X
chromosomes; males
have an X and a Y
•In mammals, X and Y
are sex chromosomes
The Sperm
Determines
the Sex of
the Progeny
More generally,
the contribution
of the
heterogametic
sex determines
the sex of the
progeny
Fig. 10-15
Sex-Linkage
• Most sex-linked traits have their loci on the X
chromosome. (Until recently it was thought Y had
•
•
no genes, but it has a few)
Females can be heterozygous for sex-linked traits
– Can mask the presence of a recessive gene.
– Hence, the ominous-sounding term “carrier”
female.
In contrast, males directly express whatever alleles
are on their X chromosomes. Both dominant and
recessive sex-linked traits are expressed, because
males are hemizygous at these loci, i.e., they have
only one X chromosome and thus only one allele.
Heterozygous
Female
X
X
A
a
Hemizygous
Male
X
Y
a
No ‘A’ locus
Males are more likely than
females to express sex-linked
traits
• Sex-biased inheritance is suggestive of sexlinkage
• Example: Red-green color blindness
– Chromosome types:
• xc Colorblind X
• xC Normal X
•Y
– Genotypes
x C xC
xC xc
xc xc
xC Y
xcY
Normal ♀
Carrier ♀
Colorblind ♀
Normal ♂
Colorblind ♂
Color Blindness
Solving genetics problems
• Select symbols for all alleles and make a
key to phenotypes if appropriate
• Determine genotypes of P generation
– True-breeding? Deduce from phenotype?
– If unknown, can it be deduced from F1?
• Determine the possible gametes for all
relevant loci
• Set up Punnett square or use probabilities
for specific genotypes or phenotypes
Example
• In spiders, the allele for hairy
abdomen is dominant to the allele for
hairless abdomen. If a hairy and a
hairless individual are crossed:
–What ratio of hairy to hairless
abdomens would you expect in their
offspring?
–What would the observed ratios tell
you about the parents?
Beyond Mendelian genetics
• Dominance explained
• Incomplete dominance
• Codominance
• Multiple alleles
• Pleiotropy
• Gene interactions
– Epistasis
– Multilocus traits
• Polygenic traits
• Genotype and environment interact to
produce phenotype
Complete Dominance
• Dominant alleles produce a functional gene
product (a structural protein or enzyme)
• Recessive alleles do not produce a functional
gene product (or produce a product lacking
something the dominant allele produces)
• When an individual possesses two recessive
alleles, no functional product is made
• But an individual with one or two dominant
alleles produces the functional product
Incomplete
Dominance
•Heterozygote is
intermediate
•Phenotype reveals
genotype
•Heterozygote
produces
intermediate amount
of gene product
(e.g., pigment)
•Often use upper
case with superscript
to distinguish alleles
Codominance
– Heterozygote displays
characteristics of two
dominant alleles
– Each allele produces a
distinct functional gene
product
– Roan horses have red AND
white hair, each coded by a
different allele
– ABO Blood type is another
example, but with multiple
alleles
Multiple alleles
• ABO blood typing: several red blood cell
antigens can occur on cell surface
• Alleles:
– IA
– IB
– io
 A antigen
 B antigen
 no antigen
• Genotypes & phenotypes
– IAIA, IAio
– IBIB, IBio
– IA IB
– ioio
=
=
=
=
Type
Type
Type
Type
A
B
AB (codominant!)
O
Pleiotropy
• One locus influences more than one trait
• Example: Sickle cell
– Influences both sickling of red blood cells in
low [O2] and resistance to malaria
Epistasis (gene interaction)
• Genotype at one locus can mask effect of
genotype at another locus
• E.g., color pattern in many animals may
be influenced by many loci, but can be
overridden if homozygous recessive (ee
or cc) at the color locus
• Enzyme needed to make melanin pigment
is not produced in ee individuals, so no
melanin-based pigments can be produced
by alleles at B locus
Fig. 6.13
Figure 10-21
Page 213
Black
BBEE, BbEE, BbEe
Yellow
BBee, Bbee, bbee
Chocolate
bbEE, bbEe
Epistasis and coat color in Labs
Multilocus traits (gene interaction)
• Two or more loci may control a single trait (e.g., combs)
• The pea gene controls the shape of the comb (’single’ vs ‘pea’)
if the individual is recessive at the rose locus
• The rose gene controls the shape of the comb (’single’ vs ’rose’
if the individual is recessive at the pea locus.
• When P and R are both present (P_R_), a walnut comb is
•
expressed.
Comb phenotype depends on genotype at both loci
• What ratio of phenotypes do you expect in
the progeny of a cross between two PpRr
walnut combed individuals?
9:3:3:1 – same as F1 dihybrid cross for
two independent traits!
Polygenic inheritance
• Multiple loci affect a single trait that varies
continuously in phenotypic value
• E.g., height, weight, skin color
• Each allele can be thought of as ‘additive’
– Contributing an incremental increase in the
phenotypic value
– Presumably by increasing amount of some
gene product produced
• Sub-discipline that studies continuous
variation is quantitative genetics
Polygenic inheritance
Example: skin color
• Assume 3 loci contribute
•
•
their products to produce
a gradual gradation of skin
color or height.
Their effects can be
thought of as additive –
each dominant allele
increases darkness or
height by one unit.
Not blending inheritance –
parental genotypes and
corresponding phenotypes
can be recovered.
Punnett square
yields expected
frequency
distribution
Still can perhaps
analyze this as a
Mendelian trait
Environment also
plays a role – tends
to blur differences
between genotypes
 more continuous
distribution
180
160
Number of men of each height
As add more loci,
distribution becomes
more continuous 
normal
distribution or
‘bell curve’
140
120
100
80
60
40
20
0
143
163
173
183
193
Height in centimeters of 1083 adult men
Phenotype reflects an interaction
between genotype and environment
• Think of genotype as a ‘recipe’
– Baking conditions, quality of ingredients,
etc., can influence final product
• Environment can alter both qualitative and
quantitative traits
– Human height or weight
– Caterpillars, Tadpole tails
Diet
Catkins
Leaves
%Catkin morphs
94 (n=18)
6 (n=18)
Phenotypic plasticity: ability of a single
genotype to produce different phenotypes in
different environments
With Dragonflies
Without Dragonflies
Mendelian Genetics
•Vocabulary
•Mendel’s experiments
•The monohybrid cross
•The dihybrid cross
•Mendel’s laws – segregation and
independent assortment
•Solving Genetics problems
•Non-Mendelian Genetics
•From genotype to phenotype
Testing for match to expected
ratios in a cross
• Compare OBSERVED number of
offspring of each phenotype to the
EXPECTED numbers if your hypothesis
is correct
• Test using Chi-square statistic
–Requires counts instead of ratios or
fractions
Chi-square
example
using
Mendel’s
dihybrid
cross of
peas
• Observe:
–
–
–
–
315 yellow smooth
108 green smooth
101 yellow wrinkled
32 green wrinkled
If Segregation and
independent assortment
hypotheses are supported
• Expect:
–
–
–
–
9/16
3/16
3/16
1/16
How can we tell if these are close enough?
Chi-square test
• Multiply expected proportions or
fractions by sample size (in this case
556) to get expected counts
–9/16 * 556 = 312.75
–3/16 * 556 = 104.25
–3/16 * 556 = 104.25
–1/16 * 556 = 34.75
Chi Square table
O - E (O - E)2
(O - E)2
E
Phenotype
Obs Exp
Yellow Smooth
315 312.75 2.25
5.0625
0.0162
Green Smooth
108 104.25 3.75
14.0625
0.1349
Yellow Wrinkled
101 104.25 -3.25 10.5625
0.1013
Green Wrinkled
32
Total
556
34.75
-2.75 7.5625
0.2176
Calculated
x2=
0.4700
Is observed x2 > critical x2 value for n-1 df, which in this case is 7.99?, if so,
result deviates significantly from expected.
df
Probability of exceeding the critical value
0.10
0.05
0.025
0.01 0.001
1
2
3
4
5
6
7
8
9
10
2.706
4.605
6.251
7.779
9.236
10.645
12.017
13.362
14.684
15.987
3.841
5.991
7.815
9.488
11.070
12.592
14.067
15.507
16.919
18.307
5.024
7.378
9.348
11.143
12.833
14.449
16.013
17.535
19.023
20.483
6.635
9.210
11.345
13.277
15.086
16.812
18.475
20.090
21.666
23.209
10.828
13.816
16.266
18.467
20.515
22.458
24.322
26.125
27.877
29.588
Note: You will not likely need more than 10 df