Chapter 2

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Chapter 02

MENDELIANMENDELIANINHERITANCE

Many theories of inheritance have been proposed to explain t i i f

INTRODUCTION

transmission of hereditary traits

The most prominent are

Theory of pangenesis Theory of preformationism Blending theory

Factors that control hereditary traits are maleable

They can blend together generation after generation

AverageOffspring

Parent 1 Parent 2

Mendel’s Laws of Inheritance

Gregor Mendel’s pioneering experiments with garden peas refuted all of the above

Mendel was successful, when his predecessors were not, because he

1. Chose a very suitable organism:

Pisum sativum

2. Took time to ensure pure breeding

3. Concentrated on one or a few discontinuous characters at a time

4. Adopted quantitative form of analysis

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Mendel’s Experimental System

Used pure-breeding lines of garden peas

Selected 7 traits to study, each of which had two distinguishable appearancesFigure 2.4

He carefully designed his experiments and gathered mathematical data

Figure 2.4

1. Cross-fertilization

Pollen and egg are derived from different plants

Mendel’s Experimental System

Refer to Figure 2.3

2. Self-fertilization

Pollen and egg are derived from the same plan

Figure 2.3

MONOHYBRID CROSSES

Mendel crossed individuals from two different lines that differ in only one traitParental generation = Pure bred lines

Fi t fili l ti Off i fFirst filial generation = Offspring of cross-fertilization of parentals

Second filial generation = Offspring of self-fertilization of F1 plants

Third filial generation = Offspring of self-fertilization of F2 plants

Figure 2.5

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Mendel also performed the reciprocal cross, where dwarf plants were pollinated using pollen from tall plants The results were the same

For all seven traits studied

1. The F1 generation showed only one of the two parental traits

2. The F2 generation showed an ~ 3:1 ratio of the two parental traits

3. Reciprocal crosses yielded similar results

P Cross F1 generation F2 generation Ratio

Tall X dwarf stem

All tall 787 tall, 277 dwarf

2.84:1

Round X wrinkled seeds

All round 5,474 round, 1,850 wrinkled

2.96:1

Yellow X Green seeds

All yellow 6,022 yellow, 2,001 green

3.01:1

DATA FROM MONOHYBRID CROSSES

Purple X white flowers

All purple 705 purple, 224 white

3.15:1

Axial X terminal flowers

All axial 651 axial, 207 terminal

3.14:1

Smooth X constricted pods

All smooth 882 smooth, 229 constricted

2.95:1

Green X yellow pods

All green 428 green, 152 yellow

2.82:1

These results refuted a blending mechanism of heredity

MONOHYBRID CROSSES

Indeed, the data suggested a particulate theory of inheritance

Mendel postulated the following

1. A pea plant contains two discrete hereditary factors, one from each parent

These factors are now referred to as alleles

2 Th t f t b id ti l 2. The two factors may be identical or different

If the alleles are identical the organism is referred to as homozygous

If the alleles are different the organism is referred to as heterozygous

3. When the two factors of a single trait are different, one is dominant and its effect can be seen, while the other is recessive and is not expressed

4. During gamete formation, the paired factors segregate randomly so that half of the gametes received one factor and half of the gametesreceived one factor and half of the gametes received the other This is Mendel’s First Law, or the Law

(Principle) of Segregation

5. The union of male and female gametes is a random process that reestablishes the paired units

Alleles are designated by letters derived from the specific characteristic they control

Usually the letter chosen is the first letter of the dominant (or recessive) trait

For example,

T = Tall plants

t = dwarf plants

Alternatively,

D = Tall plants

d = dwarf plants

Let’s introduce a few more terms

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Genotype = The specific allelic composition of an individual

With respect to seed shape, a pea plant may have different genotypes3

TT ; Tt ; tt

Phenotype = The physical appearance of anPhenotype The physical appearance of an individual

With respect to seed shape, a pea plant may have different phenotypes2

Tall ; Dwarf

Note that the phenotype is a product of the genotype and the environment

Genotype Notation

Genotype Description

Phenotype

TTHomozygous

dominantTall

Tt Heterozygous Tall

ttHomozygous

recessiveDwarf

Information about parents can be used to predict genotypes and phenotypes of offspring

Punnett square

A grid that enables one to calculate the results of simple genetics crosses

Gametes

Genotypes

Refer to p. 24

We will illustrate the Punnett square approach using the cross of heterozygous tall plants as an example

1. Write down the genotypes of both parentsM l t Tt Male parent = Tt

Female parent = Tt

2. Write down the possible gametes each parent can make. Male gametes: T or t Female gametes: T or t

3. Create an empty Punnett square

4. Fill in the Punnett square with the possible genotypes of the offspring

5. Determine the relative proportions of genotypes and phenotypes of the offspring

Genotypic ratio

TT : Tt : tt

1 : 2 : 1

Phenotypic ratio

Tall : dwarf

3 : 1

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Mendel verified his principle of segregation by performing a test-cross In a test-cross an organism with known

phenotype but unknown genotype is crossed with an organism that has the recessive phenotype (and hence is homozygous recessive)

F l if l t i t ll it ld b For example, if a pea plant is tall it could be homozygous dominant (true-breeding) or heterozygous This unknown genotype is designated

T –

A test-cross will determine its correct genotype

Tall X Dwarf

If all offspring are tall

=> the genotype is TT

If the offspring are in a ratio of 1 tall:1 dwarf

=> the genotype is Tt

To solve problems dealing with standard genetic crosses, follow these general rules:

1. Lay out the problem as a simple diagram

2 Assign letters for the alleles of the2. Assign letters for the alleles of the genes in question

3. Determine the genotypes of the parentals

4. Do a Punnett square

Application problem

In humans, free earlobes are dominant to attached ear lobes. This characteristic is under the control of a single gene.

1. Elias and his wife Nadine have attachedearlobes. What percentage of their childrenis expected to have attached earlobes?

2. Peter and his wife Aida have freeearlobes. Peter’s mom and Aida’s dad haveattached earlobes. What percentage of Peterand Aida’s children are expected to haveattached ear lobes?

DIHYBRID CROSSES

Mendel crossed individuals from two different lines that differ in two traits

Refer to Experiment 2B Refer to Experiment 2B

P Round, yellow X Wrinkled, green

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Trait 1 = Seed texture

Trait 2 = Seed color

Figure 2.8

P Cross F1 generation F2 generation

R d All d 315 d ll d

DATA FROM DIHYBRID CROSSES

Round,

Yellow seeds X wrinkled, green seeds

All round, yellow

315 round, yellow seeds

101 wrinkled, yellow seeds

108 round, green seeds

32 green, wrinkled seeds

2.97 : 1

Mendel looked at each trait in the F2 separately

3.18 : 1

These values are very close to segregation expectations

If the genes do assort independently, what will the predicted phenotypic ratio be?

9 3 3 1

Thus, Mendel proposed his second law, the law (principle) of independent assortment During gamete formation, the segregation of any

pair of hereditary determinants is independent of the segregation of other pairs

P Cross F1 generation F2 generation Ratio

Round,

Yellow seeds x Wrinkled, green seeds

All round, yellow 315 round, yellow seeds

101 wrinkled, yellow seeds

108 round, green seeds

32 green, wrinkled seeds

9.8

3.2

3.4

1.0

9:3:3:1

Figure 2.8

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Figure 2.9 Mendel verified this principle by, again,

performing a test cross

Cross F2 yellow, round plant with a

Green, Wrinkled

If the offspring are 1 yellow, round plant :

YYRr

If the offspring are 1 yellow, round plant : 1 yellow, wrinkled plant, what is the genotype of the dominant parent?

Note: Offspring of a test-cross are always in equal proportions!

Crosses that involve three pairs of traits

TRIHYBRID CROSSES P

Gametes

F1

Gametes

A Punnett square in this case will have squares! 64

Therefore it will be too cumbersome!

An alternative is the forked-line method Treat each gene separately Consider phenotypic (or genotypic) differences

In the F2 generation, 3/4 of all organisms will have the A phenotype,

and 1/4 ill ha e the a phenot peand 1/4 will have the a phenotype 3/4 of all organisms will have the B phenotype,

and 1/4 will have the b phenotype 3/4 of all organisms will have the C

phenotype, and 1/4 will have the c phenotype

Refer to Solved Problem 4 (pp. 37-38)

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linked One caveat: The genes must not be

General Rules for Multihybrid Crosses

If n = # of heterozygous gene pairs involved

Number of possible parental gametes

2n

Number of different genotypes in zygotes

3n

Number of different phenotypes in zygotes

2n

Proportion of homozygous recessive in F2

1/4n

PEDIGREE ANALYSIS

Refer to pp. 29-30 and 602-609

Determining the inheritance pattern of human traits is somewhat difficulthuman traits is somewhat difficult 1. Controlled matings are not possible

2. Sibship sizes are relatively small 3. Complete penetrance is sometimes

lacking 4. Degrees of expressivity sometimes vary

Pedigrees

Diagrams that show the relationship among members of a family, as well

PEDIGREE ANALYSIS

among members of a family, as well as their status with respect to a particular hereditary condition

Figure 2.11

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Traits that are determined by a single gene are of five main types

Autosomal dominant

Autosomal recessive Autosomal recessive

X-linked dominant

X-linked recessive

Y-linked

AUTOSOMAL DOMINANT TRAITS

1. Trait usually present in every generation 2. Members of both sexes affected equally 3. Unaffected parents produce unaffected

offspringoffspring 4. Two affected parents can produce

unaffected offspring 5. On average half the children of an

affected parent will be affected

Polydactyly

Autosomal Dominant Disorders Brachydactyly

Syndactyly Achondroplasia

Autosomal Dominant Traits

1. Trait tends to skip generations 2. Members of both sexes affected equally 3. Unaffected parents can have an affected

child

AUTOSOMAL RECESSIVE TRAITS

child 4. Two affected parents cannot have an

unaffected child 5. Traits are more likely to occur in

consanguineous marriages

Albinism

Absence or partial deficiency of melanin in the skin, eyes and hair

It occurs in a wide variety of animals

Autosomal Recessive Disorders

It occurs in a wide variety of animals

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Cystic fibrosis The most common

lethal genetic disease among caucasians

Autosomal Recessive Disorders

Mainly characterized by serious digestive and respiratory problems

Average life expectancy ~ 30 yrs

1. Trait may skip generations 2. Most affected individuals are male 3. Affected males result from affected

mothers or mothers who are carriers

X-LINKED RECESSIVE TRAITS

mothers or mothers who are carriers 4. Affected females come from affected

fathers and affected or carrier mothers 5. Affected females will have affected

sons 6. Trait can never be transmitted directly

from father to son

Red-green color blindness

X-linked Recessive Disorders

Hemophilia

1. Trait does not skip generations 2. Affected males must come from affected

mothers 3. Affected females come from affected

mothers or fathers

X-LINKED DOMINANT TRAITS

4. Affected males have no normal daughters and no affected sons

5. Affected heterozygous females transmit the trait to approximately half of their children of either sex

6. Affected homozygous females transmit the trait to all their children

Vitamin D-resistant ricketsRi k t i di d

X-linked Dominant Disorders

Faulty tooth enamel

Rickets is a disorder involving softening and weakening of the bones (of children) primarily caused by lack of vitamin D

This form of rickets is caused by lack of phosphate

1. Trait only affects males

2. Affected males get it from their fathers and give it to their sons

Y-LINKED TRAITS

Hypertrichosisof the ear

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Penetrance

The proportion of individuals with a specific genotype who manifest the corresponding phenotype

% Penetrance = # of indiv. with correct phenotype

# of individuals with genotype coding

for phenotype

Complete penetrance

Incomplete penetrance

The ABO blood group genes show almost complete penetrance

You will find out later!

Penetrance

Neurofibromatosis shows 50-80% penetrance

Polydactyly also shows incomplete penetrance

Note: Incomplete penetrance may cause a phenotype to skip a generation

Consider this pedigree that traces polydactyly in a family

Expressivity The degree to which a penetrant gene is

phenotypically expressed

Identical known genotype

A range of phenotypes

is produced

Variable Expressivity

If Neurofibromatosis penetrates, it may be exhibited in various forms

Mildest form

Café-au-lait spots

Intermediate form Intermediate form

Café-au-lait spots

Freckling

Severe form

Neurofibromas of various sizes

Speech impediments

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If polydactyly penetrates, it may be exhibited in various forms

What are the reasons for incomplete

Note: Penetrance and expressivity are used most often with autosomal dominant traits

penetrance and variable expressivity?

Gene interactionsGene interactions

Environmental effectsEnvironmental effects

Siamese cats have darker fur at their extremities

Phenylketonuria

Refer to pp. 8-9

An autosomal recessive disorder An autosomal recessive disorder

Defect is in the gene that encodes phenylalanine hydroxylase

Melanin

The metabolic pathway of phenylalanine breakdownFigure 13.1

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Affected individuals cannot metabolize phenylalanine

Phenylalanine will accumulate and be converted to phenylpyruvic acid

These are toxic and will cause mental and physical abnormalities

Newborns are now routinely screened for PKU

Individuals with the disease are put on a strict dietary regimen

Their diet is very low on phenylalanine

These individuals tend to develop normally

Phenocopy

An organism whose phenotype has been modified to resemble the phenotype of a different mutant organismg

Phocomelia

Thalidomide

Phocomelia Suppression of limb

development

Caused by a dominant allele with variable expressivityp y

Thalidomide Sedative

Between 1959-1962 children with shortened limbs were born to mothers who took the drug early in their pregnancy

PROBABILITY AND STATISTICS

Probability deals with chance or random events

The laws of probability allow us to predict the outcomes of crosses, in addition to many other processes

Probability =

# of favorable casestotal # of cases

It ranges fromg

0 to 1

If an event has a probability of p, the probability of all alternative events is q

Rules For Combining Probabilities

1. Product rule

2. Sum rule2. Sum rule

3. Binomial expansion equation (binomial theorem)

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1. Product rule

The probability of the occurrence of independent events is the product of their respective probabilities

Example: Probability of rolling a 3 ANDExample: Probability of rolling a 3 AND then a 4 with a die?

P(3 and 4) =

P(3) X P(4)

1/6 X 1/6

1/36

2. Sum rule

The probability of the occurrence of one of several mutually exclusive events is the sum of their respective probabilities

Example: Probability of rolling a 3 OR aExample: Probability of rolling a 3 OR a 4 with a die?

P(3 or 4) =

P(3) + P(4)

1/6 + 1/6

1/3

2. Sum rule Note: If the 2 events are not mutually

exclusive, then

P(A or B) = P(A) + P(B) - [P(A) X P(B)]

Example: Probability of drawing an ace OR a heart from a deck of cardsa heart from a deck of cards

P(A or ) =

P(A) + P() – [P(A) X P()]

4/52 + 13/52 – (4/52 X 13/52)

17/52 – (1/52)

16/52

3. Binomial Expansion Equation

The probability of the occurrence of unordered mutually exclusive events is defined by the binomial theorem

n!! ( )!

px qn – x P = x! (n – x)!

p q

where

p = probability that the unordered number of events will occur

n = total number of events

x = number of events in one category

p = individual probability of x

q = individual probability of the other category

Refer to the example on pp. 32-33

Question

Two heterozygous brown-eyed (Bb) yg y ( )individuals have five children

What is the probability that two of the couple’s five children will have blue eyes?

Applying the binomial expansion equation

Step 1: Calculate the individual probabilities

This can be obtained via a Punnett square

1/4 P(blue eyes) = p =

3/4 P(brown eyes) = q =

Step 2: Determine the number of events

n = total number of children = 5

x = number of blue-eyed children = 2

Step 3: Substitute the values for p, q, x, and nin the binomial expansion equation

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n!

x! (n – x)!px qn – x

P =

5!

2! (5 – 2)!(1/4)2 (3/4)5 – 2

P =

5 X 4 X 3 X 2 X 1

(2 X 1) (3 X 2 X 1)(1/16) (27/64)P =

Therefore 26% of the time, a heterozygous couple’s five children will contain two with blue eyes and three with brown eyes

(2 X 1) (3 X 2 X 1)

P = 0.26 or 26%

Use Of Rules

1. What is the probability of rolling two dice together and obtaining a 3 and a 6?

2. Consider two parents heterozygous for albinism. a) What is the probability that they will

have an albino child?

b) What is the probability that they will have 3 normal children in a row?

c) What is the probability that they will have, in order, 2 normal children, an albino child, and then a normal child?

d) What is the probability that they will have 3 normal children and an albino child, in any order?

CHI-SQUARE ANALYSIS

The 3:1 and 9:3:3:1 ratios are hypothetical ratios based on the following assumptions

Dominance/recessiveness of alleles

Complete segregation

Independent assortment

Random fertilization

Independent assortment and random fertilization are chance events

Therefore, they are subject to random

CHI-SQUARE ANALYSIS

y jfluctuations

To minimize the possibility that chance deviation will significantly alter the outcome, we need to look at very large numbers

How can we evaluate the likelihood that deviations are NOT significant, but due to chance?

The chi-square test is a statistical method e c squa e test s a stat st ca et odused to determine “goodness of fit”

This refers to how close the observed data are to those predicted from a hypothesis

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The basic steps are the following:

1. Establish a NULL hypothesis

It assumes that there is no real difference between the measured values and the predicted values

2. Apply 2 formula

(O – E)2

E

3. Interpret 2 value

This will ultimately lead to the acceptance or rejection of the null hypothesis

Consider the following example in Drosophila melanogaster (pp. 33-35)

Gene affecting wing shape

c+ = Normal wing

c = Curved wing

Gene affecting body color

e+ = Normal (gray)

e = ebony

Note:

Th ild t ll l i d i t d ith + i The wild-type allele is designated with a + sign

Recessive mutant alleles are designated with lowercase letters

The Cross:

A cross is made between two true-breeding flies (c+c+e+e+

and ccee). The flies of the F1 generation are then allowed to mate with each other to produce an F2 generation.

The outcome F1 generation

All offspring have straight wings and gray bodies

F2 generation

193 straight wings, gray bodies

69 straight wings, ebony bodies

64 curved wings, gray bodies

26 curved wings, ebony bodies

352 total flies

Applying the chi square test

Step 1: Propose a hypothesis that allows us to calculate the expected values based on Mendel’s laws

The two traits are independently assorting

Step 2: Calculate the expected values of the four phenotypes, based on the hypothesis

According to our hypothesis, there should be a 9:3:3:1 ratio on the F2 generation

Phenotype

straight wingsstraight wings, gray bodies

straight wings, ebony bodies

curved wings, gray bodies

curved wings, ebony bodies

Step 2: Calculate the expected values of the four phenotypes, based on the hypothesis

According to our hypothesis, there should be a 9:3:3:1 ratio on the F2 generation

Phenotype Expected probability

Expected number

straight wings 9/16straight wings, gray bodies

9/16

straight wings, ebony bodies

3/16

curved wings, gray bodies

3/16

curved wings, ebony bodies

1/16

Step 2: Calculate the expected values of the four phenotypes, based on the hypothesis

According to our hypothesis, there should be a 9:3:3:1 ratio on the F2 generation

Phenotype Expected probability

Expected number

straight wings 9/16 9/16 X 352 = 198straight wings, gray bodies

9/16 9/16 X 352 = 198

straight wings, ebony bodies

3/16 3/16 X 352 = 66

curved wings, gray bodies

3/16 3/16 X 352 = 66

curved wings, ebony bodies

1/16 1/16 X 352 = 22

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Step 3: Apply the chi square formula

(O1 – E1)2

E1

(O2 – E2)2

E2

(O3 – E3)2

E3

(O4 – E4)2

E4

+ + +

(193 – 198)2

198

(69 – 66)2

66

(64 – 66)2

66

(26 – 22)2

22+ + +

198 66 66 22

0.13 + 0.14 + 0.06 + 0.73

1.06

Step 4: Interpret the chi square value

The calculated chi square value can be used to obtain probabilities, or P values, from a chi square table

These probabilities allow us to determine the likelihood that the observed deviations are due to random chance alone

Low chi square values indicate a high probability that the observed deviations could be due to random chance alone

High chi square values indicate a low probability that the observed deviations are due to random chance alone

By convention, the “normal” level of significance is 5 %

If p ≥ 0.05, the deviation is considered to be statistically insignificant

=> Null hypothesis is accepted > Null hypothesis is accepted

If p < 0.05, the deviation is considered to be statistically significant

=> Null hypothesis is rejected

Before we can use the chi square table, we have to determine the degrees of freedom (df)

The df is a measure of the number of categories that are independent of each other

df = n – 1

where n = total number of categoriesg

In our experiment, there are four phenotypes/categories

Therefore, df = 4 – 1 = 3

Refer to Table 2.1

1.06

With df = 3, the chi square value of 1.06 is slightly greater than 1.005 (which corresponds to P = 0.80)

A P = 0.80 means that values equal to or greater than 1.005 are expected to occur 80% of the time based on random chance alone Thus, it is quite probable that the deviations

between the observed and expected values in this expt. can be explained by random sampling error

If the NULL hypothesis is rejected, then what???