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GENETICS

A Conceptual Approach FIFTH EDITION

Benjamin A. Pierce

CHAPTER 3

Basic Principles of Heredity

STUDY UNIT 2 3.1 Gregor Mendel Discovered the Basic

Principles of Heredity, p 42

• 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.

Mendel’s Success

Genetic terminology

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.

Consider multiple loci on one homologous chromosome

pair:

locus for characteristic 1 (G - locus)

locus for characteristic 2 (A - locus)

locus for characteristic 3 (B - locus)

Homozygotic: two alleles at a particular

locus are identical, e.g. BB.

Heterozygotic: two alleles at a particular

locus differ, e.g. Gg and Aa.

g G

A a

B B

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• 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.

3.2 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

Experimental

strategy

Observed results

Results

Conclusion

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.

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Class work

Apply modern genetic terminology and symbols to Mendel’s monohybrid crosses.

What Monohybrid Crosses Reveal

• 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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Chromosome theory of heredity.

Relating Genetic Crosses to Meiosis

Mendel’s principles work because they are based on the behaviour of actual chromosomes in meiosis.

The Molecular Nature of Alleles

• 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.

The Punnett square

Predicitng the Outcome of Genetic Crosses 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 short-

haired ♀ 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 (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).

Probability as a Tool in Genetics

The multiplication rule

The probability of two or more independent events occuring together is calculated by multiplying their independent probabilities (“and” is used).

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.

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.

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

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.

Class work

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.

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)?

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).

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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(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

If order of events is specified, use multiplication rule

(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

If no order is specified, requirement can be met

in many different ways

Binomial: (p + q)n = 1

p, q = respective probabilities of 2 alternative outcomes

n = number of times event occurs

n Binomial Expanded binomial

1 (p + q)1 p + q

2 (p + q)2 p2 + 2pq + q2

3 (p + q)3 p3 + 3p2q + 3pq2 + q3

4 (p + q)4 p4 + 4p3q + 6p2q2 + 4pq3 + q4

5 (p + q)5 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

Numerical coefficients for each term are determined

by using Pascal’s triangle.

In our five child family:

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

(3) 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

st 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

st

= 5! (¼)3 (¾)2

3! 2!

= 5.4.3.2.1 (¼)3 (¾)2

3. 2.1.2.1

= 0.088

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?

Class work

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

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.

When you consider multiple loci, every gamete

receives one allele from EVERY locus

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

Class work

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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 ...…

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.

Class work

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.

• 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?

3.4 Observed Ratios of Progeny May Deviate

from Expected Ratios by Chance, p 61

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.

• 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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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.

Class work