Download STUDY UNIT 2 MENDELIAN GENETICS

3.1 Mendelian / Transmission genetics
STUDY UNIT 2
How do we determine the inheritance patterns of traits?
MENDELIAN GENETICS
Gregor Johann Mendel 1866 - reported rules that
explain the inheritance of traits from parents to progeny.
Klug, Cummings & Spencer Chapter 3
Good experimental design:
Pisum sativum, easy to cultivate and cross-breed.
Seven characters, each with two contrasting forms.
Experimental methodology.
Accurate quantitative records.
Scientific analysis.
3.2 Monohybrid cross reveals how ONE trait is
transmitted from generation to generation
• Choose true breeding parents (P).
Mendel’s results when considering one trait:
Plant size:
P:
• Cross parents with phenotypic differences in one trait
e.g.
tall plants x dwarf plants.
F1:
(trait = plant size)
• Obtain monohybrid progeny (F1).
• Self-fertilize F1 to obtain next generation (F2).
F2:
• Evaluate progeny phenotypes and numbers.
• Propose genetic model.
RECIPROCAL CROSS
Mendel’s results for seven traits:
same results
Mendel’s first three postulates:
1. Unit factors in pairs
Fig 3.1
2. Dominance / recessiveness
3. Segregation
particulate unit factors determine each trait
Explanation for results of monohybrid crosses.
1
Mendel in modern genetic terminology:
Phenotype:
Physical expression of a trait.
Convention for naming genes:
Genotype:
Genetic makeup of an individual.
Trait
Genes:
Genetic factors (units of inheritance) that
determine specific traits.
Alleles:
Alternative forms of a single gene.
E.g.
Size
dominant form
recessive form
first letter of word
tall (dominant)
dwarf (recessive)
dwarf
uppercase
lowercase
D-allele
d-allele
Homozygote: Identical alleles for a specific gene.
Heterozygote: Different alleles for a specific gene.
Combining Mendel’s postulates and modern terminology:
The monohybrid cross
Fig 3.2
This type of allelic interaction = (Complete) DOMINANCE
Dominant phenotype: phenotype expressed when genotype
contains 1 or 2 copies of dominant allele
Dominant allele: allele expressed phenotypically in
heterozygote, or allele of which one copy is sufficient to
determine dominant phenotype
Punnett squares:
Fig 3.3
Recessive phenotype: phenotype expressed in absence of
dominant alleles, when genotype contains only recessive
alleles
Recessive allele: allele expressed phenotypically only in
absence of dominant allele
In this example:
The testcross: one character
F2 plant with tall phenotype
is genotype DD or Dd?
Testcross:
Dominant phenotype =
Fig 3.4
Dominant allele =
Recessive phenotype =
Recessive allele =
2
(Prac book 2.5)
Examples:
(Prac book 2.2)
In foxes, silver coat colour is governed by a recessive
allele (b) and red by the dominant allele (B). Give the
genotypic and phenotypic ratios of the progeny of:
(a)
(b)
carrier red x silver
pure red x silver
3.3 Mendel’s dihybrid cross generated a
unique F2 ratio
Dihybrid / two-factor cross involves two pairs of
contrasting traits.
Fig 3.5
In rabbits short hair is due to a dominant allele (A) and
long hair to its recessive allele (a). A cross between a
short-haired and a long-haired produces 1 longhaired and 7 short-haired rabbits.
a) Give the genotypes of the parents.
b) What was the expected phenotypic ratio of the
progeny?
c) How many rabbits were expected to have long hair?
d) Explain why only one with long hair was born.
Consider each trait separately:
Colour
Shape
F1:
all yellow
all round
F2:
9/16 + 3/16 = 12/16 yellow
12/16 round
3/16 + 1/16 = 4/16 green
4/16 wrinkled
3 yellow : 1 green
3 round : 1 wrinkled
Two pairs of contrasting traits are
Can predict frequencies of F2 phenotypes
Fig 3.6
Mendel’s fourth postulates:
4. Independent assortment
The dihybrid cross:
Fig 3.7
REMEMBER:
Rule of segregation: 2 alleles of a locus segregate, so
that 1 allele of that locus is present in a gamete.
However: every gamete receives one allele from every
locus.
E.g. for AaBb
A-allele has an equal chance to combine
with the B- or the b-allele, etc.
See p 48: A molecular explanation how Mendel’s peas
became wrinkled
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The testcross: two characters
F2 plant with dominant (yellow, round) phenotype
what is its genotype?
Drosophila nomenclature for allele symbols:
One characteristic – two phenotypes
Wild type (common)
Testcross:
Mutant (rare)
use description of mutant phenotype
to choose letter(s) to describe this gene
Fig 3.8
capital letter if mutant allele dominant
small letter if mutant allele recessive
same letter with + as
superscript for wild
type allele
E.g.
• Eye colour: red or white
•
potential allele symbols:
•
Wild type (common) = red, mutant (scarce) = white
→
Determine dominance (from crosses)
→
→
White (mutant) allele =
Red (wild type) allele =
•
•
3.4 The trihybrid cross: Mendel’s principles
apply to inheritance of multiple traits
Example:
(Prac book 2.12)
In Drosophila, ebony body colour is determined by a
recessive allele (e) and wild-type grey colour by the
dominant allele (e+). Vestigial wings are determined by a
recessive allele (vg) and wild type normal wings by the
dominant allele (vg+). Wild-type, dihybrid flies are mated
and produce 256 progeny. How many flies are expected
in each phenotypic class?
Fig 3.9
Use a Punnett square to calculate progeny ratios:
• Consider inheritance of 3 or more loci simultaneously.
• 2 alleles / locus in population, with dominance.
Remember during gamete formation:
•
•
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Alternative approach: Use the forked-line or
branch diagram method to predict ratios:
• Consider each trait separately
• Combine results
• True only if genes assort independently
Determining gametes of multihybrids: Forked-line method
E.g. Individual AaBBCcDdeeFf How many gametes?
C
A
B
c
Fig 3.10
C
a
B
c
D
e
F
f
ABCDeF
d
e
F
f
ABCdef
D
e
d
e
F
f
F
f
D
e
F
f
d
e
F
f
D
e
F
f
e
F
f
d
n heterozygous loci
General rules:
etc.
2n different gametes
Example:
Consider an individual with genotype JjMMNn.
If self-fertilisation occurs, what is the expected
genotypic ratio of the progeny?
Table 3.1
Approach 1:
Determine gametes, use Punnet square, determine
progeny genotypes and ratios.
Approach 2:
Consider each locus independently, combine
genotypic ratios using forked-line approach.
3.5 Mendel’s work was rediscovered in the early 20th
century
Unit factors, genes and homologous chromosomes
1866: Publication of Mendel’s results.
Darwin, natural selection, continuous variation.
3.6 Mendel’s postulates correlate with behaviour of
chromosomes
Fig 3.11
1879: Chromosomes.
Early 1900’s – hybridization experiments with plants.
1902: Chromosome behaviour during meiosis
Mendelian principles of segregation and independent
assortment.
→ Chromosomal theory of heredity.
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Pedigrees reveal patterns of inheritance in humans
Ensure that you can integrate and understand the
following terms and concepts with respect to mitosis,
meiosis, and Mendelian genetics:
Diploid number, haploid number, gamete, zygote,
chromosome, chromatid, homologous pair, maternal
parent, paternal parent, locus, gene , allele, segregation,
independent assortment, genotype, phenotype.
•
Designed crosses not possible in humans
•
Few offspring, long generation times
•
Construct a family tree, indicate presence or absence of
trait in question for each member
•
Pedigree analysis
Pedigree
conventions:
Pedigree analysis
Pedigree for an autosomal recessive trait, e.g. albinism
Define allelic interaction, genotypes, phenotypes:
Fig 3.14a
Example:
(Klug Ch 3 Q 26 p 67)
Consider the following pedigree.
Fig 3.13
Pedigree for an autosomal dominant trait, e.g.
brachydactyly or Huntington’s disease
Define allelic interaction, genotypes, phenotypes:
Fig 3.14b
Table 3.4
3.8 Laws of probability help to explain genetic
events
Genetic ratios are most properly expressed as
probabilities, that predict the outcome of fertilization
events.
Probabilities range from 0 (event certain not to occur) to 1
(event certain to occur).
The symbol P( ) is used to indicate probability.
P (xyz) = probability to obtain xyz.
Predict the mode of inheritance and the most probable
genotypes of each individual. Assume that the alleles A and a
control the expression of the trait.
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Analysis can be based on calculating frequencies.
Frequency
Suppose
The sum law (“either / or” law)
If two events are mutually exclusive (when “or” is
10 000 male students at Tuks
9 000 female students at Tuks
19 000 TOTAL
Frequency (male):
Frequency (female):
used), the P of one or the other of the two events
occurring is the sum of their individual P.
E.g. a dice has 6 surfaces, numbered 1 to 6, each surface
has an equal chance of landing face up.
Probability that the next student you meet will be a female
=
NB: Probabilities of all options for a single event = 1
∴ 0.526 + 0.474 = 1
Genetic example:
What is the probability of rolling either a 1 or a 4 in the
next throw?
•
If you use the word OR, the SUM law applies.
•
Summate the respective probabilities.
Consider the cross Aa x Aa. What is the probability
that any of the progeny will show the dominant trait?
(I.e. what is the P for the AA or Aa genotypes in the
progeny of this cross?)
∴P (rolling a 1 or 4) = P(1) + P(4)
=
The product law (“and” law)
When two events are independent of one another (“and”
is used), the P that they will occur together is the product
of their individual P.
E.g. a coin is tossed.
P(head) = ½ = P(tail).
What is the probability of obtaining a head on a 20c
coin and a tail on a 50c coin if you toss them together?
• If you use the word AND, the PRODUCT law applies.
• Multiply the respective probabilities.
∴ P (head on 20c) and P (tail on 50c)
=
If same coin is tossed again, or another coin is tossed,
the P remains ½ for any of the events, as they are
independent and do not influence each other.
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Genetic example:
Sum- and product laws are often combined:
Consider the cross PpRr x Pprr. What is probability
of obtaining a pprr homozygote in the progeny?
Again consider two coins that are tossed simultaneously.
What is P of obtaining heads on the one coin and tails on
the other?
There are now two possible outcomes, i.e. heads on 20c
and tails on 50c or tails on 20c and heads on 50c.
∴P [ H on 20c and T on 50c] or P [T on 20c and H on 50c]
=
Product- & sum rules may be extended to more than 2
events:
E.g.
Deck of cards has 13 types of cards (aces, twos,
threes, …kings). P to draw a particular card = 1/13.
• What is P of drawing a face card (jack or queen or king) on
your first draw?
• What is the P of drawing first an ace, and then a two, and
then a six?
Pc = P (outcome of interest) / P (specific condition)
Conditional probability (Pc):
If the probability of an outcome is dependent on a specific
condition related to that outcome.
E.g. In the F2 of a monohybrid cross involving tall and dwarf
plants, what is the probability that a tall plant is heterozygous
(and not homozygous)?
Consider only tall plants, as all dwarf plant are homozygous
recessive. Some tall plants are heterozygous.
As outcome and condition are NOT independent, we CAN
NOT use the product law.
Binomial theorem
To determine probability in the following situations:
•
•
•
Applications in genetic counselling:
If one child in a family is affected with a recessive
disorder, what is the probability that normal siblings
are carriers of the disease allele (heterozygotic)?
•
•
There are a number of occurrences of a specific
event (e.g. children born in a family),
there are two possible outcomes (boy or girl),
events must be mutually exclusive (only one or
the other can occur in individual),
events are independent (eg. birth of one child
does not influence next pregnancy),
no order is specified
(if order is specified: simply use product rule).
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E.g. In a family with 5 children, there are 6 possible
combinations of males and females:
Binomial theorem: (a + b)n = 1
a, b = respective probabilities of 2 alternative outcomes
n = number of trials
all 5 females
4 females + 1 male
3 females + 2 males
2 females + 3 males
1 female + 4 males
all 5 males
n
The children may be born in any order.
Consider option , then order of birth may be:
M, F, F, M, F or F, F, F, M, M or M, M, F, F, F ……..
or seven other possible orders.
Binomial
1
2
3
(a +
(a + b)2
(a + b)3
4
(a + b)4
etc
Expanded binomial
b)1
How do we determine the probabilities of these different
alternatives?
Exponents are determined by using pattern:
(a + b)n = an, an-1 b, an-2 b2, an-2 b3, …….., bn
Eg
numbers of events n = 6
(a + b)6
Numerical coefficients are determined by using
Pascal’s triangle:
1
n = 6:
Table 3.2
6
15
1
coefficient where
6
coefficient where
15
coefficient where
1
coefficient where
20
15
6
1
=
The probability of particular outcome can also be
calculated using the following formula:
P =
n!! as bt
s!!t!!
n = total number of events
s = number of times outcome a occurs
t
NB:
s+t=n
a+b=1
0! = 1
x0 = 1
= number of times outcome b occurs
a = probability of one outcome
b = probability of other outcome
n! / (s!t!) gives the numerical coefficient for any set of
Ex 1. What is the probability that 3 females and 2 males
will be born in families with 5 children?
n
s
t
a
b
=
=
=
=
=
Sequence of events NOT specified
expansion
use binomial
Approach 1: Pascal’s triangle
Approach 2: Formula
exponents.
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Aa
Ex 2. A husband and wife, both carriers of a recessive
genetic disorder, are planning 6 children.
x
Aa
GR: ¼ AA : ½ Aa : ¼
PR:
a. What is the probability that their first two children
will be normal, and the next four affected?
¾ normal
aa
: ¼ affected
a =
b =
b. What is the probability that only one of their children
will be affected?
a. Order of events specified - use product law.
b. Order of events not specified - use Pascal’s
triangle or formula.
n =
s =
t =
3.9 Chi-square analysis evaluates the influence
of chance on genetic data
• Deviations from expected ratios are often observed.
• Due to random fluctuations of chance events.
• Error much greater with small samples.
• How do you evaluate observed deviation?
E.g. In a monohybrid cross Aa (purple) x Aa (purple) you
would expect progeny in the ratio 3 purple: 1 white.
Null hypothesis (H0): Assume data will fit a 3:1 ratio.
If 5 progeny individuals are analysed
be purple.
by chance all may
If 1000 progeny individuals are analysed
more accurate
data expected.
Expected number
750 purple : 250 white
You may observe
690 purple : 310 white
The observed deviation from an expected ratio may be:
• attributed to chance alone (sampling error)
•
not attributed to chance alone (probably other
reasons)
Statistical test for goodness of fit of H0
chi-square (χ
χ2) analysis
χ2 = statistical test to determine if
Is this deviation acceptable to still fit a 3:1 ratio?
Will the H0 be rejected or not rejected?
10
1.
Formulate hypothesis.
How do we decide to reject or not reject a hypothesis?
2. Calculate χ2 value from results.
χ2 =
(observed - expected) 2
expected
3. Interpret χ2 value.
If deviation is small
small χ2 value
If deviation is large
bigger χ2 value
Compare calculated χ2 value from the data with a critical
χ2 value from a standard chi-square table or graph.
Fig 3.12
To determine the critical χ2 table value you need two
reference points:
Note:
Fig 3-12 in the 9th (new) edition of the textbook is
correct.
Please correct the legend of Fig 3-12 in the 8th (old)
edition of Klug p 55 so that it reads as follows:
(a) …….. (b) Table of χ2 values for selected values of
df and p. χ2 values that lead to a p value of 0.05 or
greater (darker blue ares) justify failure to reject the
null hypothesis. Values leading to a p value of less
than 0.05 (lighter blue areas justify rejecting the null
hypothesis. For example ……
1. Degrees of freedom (df)
Df =
Need to be taken into account as with more categories
more deviation is expected as a result of chance.
2. Probability
Standard for biological samples usually p value of 0.05.
p < 0.05: Probability < 5% (very small) that
p > 0.05: Probability > 5% that
•
E.g. the critical χ2 table value for df = 5 and p = 0.05 is 11.07
If calculated χ2 value < tabulated value
Deviation not significant , and probably due to
chance.
Fig 3.12 (b)
•
If calculated χ2 value > tabulated value
Observed ratios differ significantly from the expected
ratios, the deviation can not be explained by chance
alone.
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Example:
(Klug Ch 3 Q 23 p 67)
In one of Mendel’s dihybrid crosses, he observed 315
round yellow, 108 round green, 101 wrinkled yellow, and
32 wrinkled green F2 plants. Analyse these data using a
χ2 test to see if they fit a 9:3:3:1 ratio.
1.
2.
3.
Formulate hypothesis.
Calculate χ2 value for data.
Interpret χ2 value.
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