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STUDY UNIT 2
3.1 Gregor Mendel Discovered the Basic
Principles of Heredity, p 42
Benjamin A. Pierce
Mendel’s Success
GENETICS
A Conceptual Approach
FIFTH EDITION
CHAPTER 3
Basic Principles of Heredity
• Choice of experimental subject, Pisum sativum, the
pea plant - easy to cultivate, short generation time,
produces many offspring (seeds), large number of
pure breeding varieties, chose seven characters each
with only two contrasting forms.
• Good experimental methodology, accurate record
keeping.
• Interpreted results by using mathematics.
• Formulated and tested hypotheses.
Genetic terminology
Consider multiple loci on one homologous chromosome
pair:
locus for characteristic 1 (G - locus)
g
G
locus for characteristic 2 (A - locus)
A
a
B
B
locus for characteristic 3 (B - locus)
Homozygotic: two alleles at a particular
locus are identical, e.g. BB.
This individual is heterozygous for the R-locus, with genotype Rr.
Other individuals of this species could also have genotype Rr, or
could be homozygous RR or homozygous rr for the R-locus.
Heterozygotic: two alleles at a particular
locus differ, e.g. Gg and Aa.
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3.2
• An individual inherits only the alleles of the genotype.
• The phenotype is determined by the genotype (the
interaction of both alleles at a given locus) PLUS
environmental factors (that can have a small or large
effect depending on the trait).
• Mendel carefully observed phenotypes, this allowed
him to deduce genotypes and the rules governing
their inheritance.
Monohybrid Crosses Reveal the Principle of
Segregation and the Concept of Dominance,
p 45
Monohybrid cross: cross between two parents that
differ in a single characteristic
e.g. ♂ round seeds x ♀ wrinkled seeds
In the reciprocal monohybrid cross, the parents
have the opposite phenotypes
e.g. ♂ wrinkled seeds x ♀ round seeds
NOTE Mendel’s application of the scientific method.
Question
Observed results
Experimental
strategy
Results
Mendel’s conclusions based on monohybrid crosses:
• Conclusion 1: One character is encoded by two
genetic factors (two alleles present for a gene).
• Conclusion 2: The two genetic factors (alleles)
separate when gametes are formed, one allele
goes into each gamete.
• Conclusion 3: The concept of dominant and
recessive traits.
• Conclusion 4: Two alleles separate with equal
probability into the gametes.
Conclusion
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Class work
What Monohybrid Crosses Reveal
Apply modern genetic terminology and symbols to
Mendel’s monohybrid crosses.
• Principle of segregation: (Mendel’s first law)
Each individual diploid organism possesses two
alleles for any particular characteristic. These two
alleles segregate when gametes are formed, and
one allele goes into each gamete.
• The concept of dominance:
When two different alleles are present in a
genotype, only the trait encoded by one of them the “dominant” allele - is observed in the phenotype.
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Relating Genetic Crosses to Meiosis
Chromosome theory of heredity.
The Molecular Nature of Alleles
Mendel’s principles work because they are based on
the behaviour of actual chromosomes in meiosis.
Predicitng the Outcome of Genetic Crosses
The Punnett square
• Remember: alleles are different variations of the DNA
sequence of a specific gene.
• How does an allele determine a phenotype?
• Consider R/r alleles for round/wrinkled pea seed
shape.
• R-locus on chromosome 5 encodes enzyme SBEI.
R-allele: normal functional enzyme
r-allele: mutation in DNA, nonactive enzyme.
• RR and Rr genotypes: sufficient levels of normal
enzyme, starch converted as required, water balance
normal, round seed.
rr genotype: only nonfunctional enzyme present,
starch not converted, water balance in seed disrupted,
wrinkled phenotype.
Examples:
1. In foxes, silver coat colour is governed by a recessive allele
(r) and red by the dominant allele (R). Give the expected
genotypic and phenotypic ratios of the progeny of:
a. Carrier red x silver.
b. Pure red x silver.
2. In rabbits short hair is due to a dominant allele (L) and long
hair to its recessive allele (l). A cross between a shorthaired ♀ and a long-haired ♂ produces 1 long-haired and 5
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.
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Class work
Probability as a Tool in Genetics
Probability (P) expresses the likelihood of occurrence
of an event.
P = number of times a particular event occurs
number of all possible outcomes
P(K♥ from deck of cards) = 1/52
P(Ace) = 4/52
P can be expressed as fraction or decimal.
Probability can be determined based on:
- knowing how an event occurs (nature of event)
- knowing how often an event occurs (frequency of
event).
The multiplication rule
The probability of two or more independent events
occuring together is calculated by multiplying their
independent probabilities (“and” is used).
E.g. dice have 6 surfaces, numbered 1 to 6, each
surface has an equal chance of landing face up.
 P (rolling a 1) = 1/6 , P (rolling a 2) = 1/6 ,
and P(3) = P(4) = P(5) = P(6) = 1/6.
What is the probability of rolling a die twice, and
obtaining first a four and then a six?
• If you use the word and, the multiplication rule
applies.
The addition rule
The probability of any one of two or more mutually
exclusive events is calculated by adding these
probabilites (“or” is used).
What is the probability of rolling a die once, and
obtaining either a four or a six?
• If you use the word or, the addition rule applies.
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The application of probability to genetic crosses
Class work
The multiplication and addition rules of probability can be used
in place of the Punnett square to predict ratios of genetic
crosses.
• Consider the cross Aa x Aa. What is the probability of
obtaining AA progeny?
• Consider the cross Aa x Aa. What is the probability that any
of the progeny will show the dominant phenotype (A_)?
(i.e. what is the probability for the AA or Aa or aA genotypes in
the progeny of this cross?)
Punnett square usually easier for simple monohybrid crosses.
Probability usually easier for complex multi-locus crosses.
Conditional probability
Sometimes extra information is available that influences
the calculation of probability.
Eg. Cross two heterozygous tall peas (Tt x Tt).
What is the probability for a tall plant in the progeny to
be heterozygous (i.e. not homozygous)?
Tt x Tt →
¼ AA : ½ Aa : ¼ aa
(of 1:2:1)
Caanot say it is ½ from the above, as we also have
information available about the phenotype.
Consider only tall plants, then the expected ratio is
1 homozygous (TT) : 2 heterozygous (Tt).
Thus probability for a tall plant to be heterozygous is ⅔.
The binomial expansion and probability
The binomial expansion is used to determine
probability in the following situations:
• There are a number of occurrences of a specific
event (e.g. children born in a family),
• there are two possible outcomes (e.g. diseased or
normal),
• events must be mutually exclusive (only one or the
other can occur in individual),
• events are independent (e.g. birth of one child does
not influence next pregnancy),
• order of events is not specified
(if order is specified: simply use product rule).
Application of conditional probability often in genetic
counselling:
If one child in a family is affected with a recessive
disorder (e.g. cystic fibrosis), what is the probability
that other normal children in this family (brothers &
sisters of ill child) are carriers of the disease allele
(heterozygous)?
Example: Two parents are heterozygous for albinism,
a recessive condition in humans
Aa x Aa → ¼ AA : ½ Aa : ¼ aa
P(normal child AA or Aa) = ¾
P(albino child aa) = ¼
If these parents want to have five children, what is the
probability for:
(1) Five albino children.
(2) First child normal, next three albino, last one
normal.
(3) Any three of the children albino, and two of the
children normal.
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If order of events is specified, use multiplication rule
(1) P (5 albino children)
= P (Alb and Alb and Alb and Alb and Alb)
= (¼)5
(2) P (First child normal, next three albino, last one
normal)
= P (Norm and Alb and Alb and Alb and Norm)
= ¾ . ¼. ¼. ¼. ¾
= (¾)2 (¼)3
Binomial: (p + q)n = 1
p, q = respective probabilities of 2 alternative outcomes
n = number of times event occurs
n
Binomial
1
2
3
(p + q)1
(p + q)2
(p + q)3
p+q
p2 + 2pq + q2
p3 + 3p2q + 3pq2 + q3
4
(p + q)4
p4 + 4p3q + 6p2q2 + 4pq3 + q4
5
q)5
(p +
If no order is specified, requirement can be met
in many different ways
(3) P (Any three of the children albino and two of the
children normal)
= P [(Alb and Norm and Norm and Alb and Alb)
or (Norm and Alb and Alb and Norm and Alb)
or (Alb and Norm and Alb and Norm and Alb)
or ……….
combine multiplication and addition rules
BUT: if there are many possible combinations,
binomial expansion is easier
Numerical coefficients for each term are determined
by using Pascal’s triangle.
Expanded binomial
p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5
etc.
Exponents are determined by using pattern:
(p + q)n = pn, pn-1 q, pn-2 q2, pn-2 q3, …….., qn
In our five child family:
(3)
n = 5
Let:
p = probability for child with albinism (¼)
q = probability for normal child (¾)
(p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5
probability
for five
albino
children
probability
for four
albino and
one normal
child
.. etc ..
probability
for one
albino and
four normal
children
P (Any three of the children albino, and two of
the children normal)
n = 5
p = probability for child with albinism (¼)
q = probability for normal child (¾)
(p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5
use this term, as this represents
probability for THREE albino (p) and TWO normal (q)
Then P = 10p3q2
= 10 (¼)3 (¾)2
= 0.088
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Alternatively calculate the probability for any
particular combination of events using the following
formula:
NB:
P = n ps qt
s+t=n
st
p+q=1
0! = 1
x0 = 1
n = total number of events
s = number of times outcome X occurs
t = number of times outcome Y occurs
p = probability of outcome X
q = probability of outcome Y
n! / (s!t!) gives the numerical coefficient for any set
of exponents.
(3) P (Any three of the children albino, and two of
the children normal)
n = 5, s = 3, t = 2
p = ¼, q = ¾
P =
=
n ps qt
st
5! (¼)3 (¾)2
3! 2!
= 5.4.3.2.1 (¼)3 (¾)2
3. 2.1.2.1
= 0.088
Class work
Example:
A newly married couple are planning to have six
children.
a. What is the probability that their first two children
will be girls, and the next four boys?
b. What is the probability that only one of their
children will be a girl?
The Testcross
Cross of an individual (often with unknown genotype)
with another individual with a homozygous recessive
genotype for the trait in question.
You are given a pea plant with a tall phenotype
→ is its genotype TT or Tt?
Testcross:
TT x tt → All Tt
Tt x tt → ½ Tt : ½ tt
Acceptable Genetic Symbols for
Alleles of a Gene
• Dominant allele (uppercase), recessive allele
(lowercase) (A, a).
• Multiple letters per allele (Hl, hl) (Azh, azh).
• Wild type or common allele (letter(s) with plus
superscript), mutant or rare allele (same letter(s)
without plus) (ye+, ye).
• Wild type allele substituted with plus sign (+, ye).
• Superscripts and subscripts (Lfr1, Lfr2).
• Slash distinguishes two allele in a genotype
(El+/ElR or +/ElR).
• Spaces separate genotypes for multiple loci
(El+/ElR G/g).
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Ratios in Simple Crosses
3.3 Dihybrid Crosses Reveal the Principle of
Independent Assortment, p 56
Dihybrid cross: cross between two parents that differ in
two characteristics, e.g.
P: ♂ round, yellow seeds x ♀ wrinkled, green seeds
F1:
All round, yellow seeds
F2: 315 round : 101 wrinkled : 108 round : 32 wrinkled
yellow
yellow
green
green
approximately 9 : 3 : 3 : 1 ratio
When you consider multiple loci, every gamete
receives one allele from EVERY locus
The Principle of Independent Assortment
Mendel carried out multiple dihybrid crosses
→ always obtained 9:3:3:1 F2 ratio.
Principle of independent assortment : (Mendel’s
second law)
The alleles at different loci separate independently of
each other.
REMEMBER:
Principle of segregation still applies: 2 alleles of a
locus segregate, so that 1 allele of that locus is present
in a gamete.
Class work
E.g. for RrYy → R-allele has an equal chance to
combine with the Y- or the y-allele, etc.
→ gametes 1 RY : 1 Ry : 1 rY : 1 ry in equal
ratios.
When these 4 types of gametes combine:
→ 9/16 round yellow , 3/16 wrinkled yellow, 3/16 round
green, 1/16 wrinkled green
= 9:3:3:1 phenotypic ratio
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Relating the Principle of Independent Assortment
to Meiosis
• Independent assortment applies to characters
encoded by loci on different chromosomes,
because
• each pair of homologous chromosomes separates
independently of all other pairs at anaphase I of
meiosis.
• Genes on the same chromosome do not assort
independently, unless they are so far apart that
crossing over will always take place between them
(see later Chapter 8).
Example:
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 of
the phenotypic class?
Class work
Applying Probability and the Branch Diagram
to Dihybrid Crosses
Consider the dihybrid (Rr Yy x Rr Yy) cross as two
monohybrid crosses:
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Now combine these monohybrid ratios by using the
multiplication rule, to obtain the ratios of the progeny
with different combinations of seed shape and colour.
When considering multiple loci, using probability rules
is much faster than using a Punnett square.
From the cross:
Aa Bb cc Dd Ee x Aa Bb Cc dd Ee
What is the probability for an offspring with the
genotype aa bb cc dd ee?
Consider each locus individually:
Aa x Aa → ¼ AA : ½ Aa : ¼ aa, so P(aa) = ¼
Bb x Bb …. etc
P (aa bb cc dd ee) = ¼ x ...…
Class work
Example:
Consider an individual with genotype Qq RR Tt.
If self-fertilisation occurs, what is the expected
genotypic ratio of the progeny?
Approach 1:
Determine gametes, use Punnet square, determine
progeny genotypes and ratios.
Approach 2:
Consider each locus independently, combine
genotypic ratios using a branch diagram.
The Dihybrid Testcross
Cross with another individual that is homozygous
recessive for both loci.
RrYy x rryy
Do using a Punnett square.
Do using a branch diagram.
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Class work
NB: Go through the Worked Problem on Pierce p 62
[p 57] in your own time.
The Goodness-of-Fit Chi-Square Test
•
•
•
•
•
E.g. in cockroaches, brown (Y) is dominant over yellow (y)
body colour.
From the cross Yy (brown) x yy (yellow)
expected progeny ratio is 1 Yy (brown) : 1 yy (yellow).
You would expect 20 brown & 20 yellow from 40 offspring.
If you observe:
22 brown & 18 yellow – probably close enough
25 brown & 15 yellow – start to doubt if this fits 1:1 ratio
5 brown & 35 yellow – would be quite sure this does not fit
expected ratio.
Require a way of evaluating how likely it is that differences we
observe are due to chance only.
3.4 Observed Ratios of Progeny May Deviate
from Expected Ratios by Chance, p 61
• For genetic crosses, certain genotypic and
phenotypic ratios are expected based on
Mendelian principles of segregation, independent
assortment and dominance.
• In experimental results, deviations from expected
ratios are often observed.
• This may be due to random fluctuations of chance
events.
• Error often greater with small samples.
• How do you evaluate the observed deviation?
• How well do observed values fit expected values?
Chi-square test: (2)
• Indicates the probability that the differences
between the observed and expected values is due
to chance.
• When the calculated probability is high, we assume
the differences are due to chance alone (null
hypothesis is true).
• When the calculated probability is low, we assume
the some significant factors other than chance are
responsible for the deviation (null hypothesis is
false).
Doing a Chi-square test: (2-test)
• Determine the expected progeny ratio.
• Note the observed results (numbers).
• Calculate the expected numbers in each class.
• Calculate the 2 value
2 =  (observed - expected) 2 =  ( o - e ) 2
expected
e
• Determine the degrees of freedom
Df = number of classes – 1
• Determine the probability associated with this
calculated 2 value at the correct degrees of freedom
from a chi-square table.
• Interpret the probability value, using 0.05 probability
as cutoff value.
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Class work
Example:
In one of Mendel’s dihybrid crosses, he observed 315
round yellow, 108 round green, 101 wrinkled yellow,
and 32 wrinkled green seeds in F2 pea plants.
Analyse these data using a 2 test to see if they fit a
9:3:3:1 ratio.
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