Mendelian Genetics. Mendels Experiments Gregor Mendel Joined the Augustinian Monastery at the age of 21 Taught in a secondary school, was fascinated with.

Mendelian Genetics

Mendel’s Experiments

Gregor Mendel Joined the Augustinian Monastery at the age of

21 Taught in a secondary school, was fascinated with

science and nature (physics, evolution, botany and natural sciences)

Attended the University of Vienna to study physics and biology

Returned to the monastery and began his experiments with the common garden plant (Pisum sativum)

Mendel’s Experimental, Quantitative Approach

• Advantages of pea plants for genetic study: Many varieties with distinct heritable features, or

characters (such as flower color); character variants (such as purple or white flowers) are called traits

Mating of plants can be controlled

Each pea plant has sperm-producing organs (stamens) and egg-producing organs (carpels)

Cross-pollination (fertilization between different plants) can be achieved by dusting one plant with pollen from another

Fig. 14-2

TECHNIQUE

RESULTS

Parentalgeneration(P) Stamens

Carpel

1

2

3

4

Firstfilialgener-ationoffspring(F1)

5

Mendel’s Experiments

Mendel’s Experimental Design Crossed only peas with desired traits, two methods

Self-fertilization – pollen from anther falls onto the stigma of the same flower before it opens

Cross-fertilization – pollen from one plant fertilizes another Mendel opened the keel and removed the anther

before self-fertilization could occur Selected 7 discrete, nonoverlapping characteristics

Flower color – purple or white

Table 14-1

Mendel’s Experiments

Mendel’s Experimental Design Grew the plants for two years to identify

homogenous, pure-breeding characteristics Started by crossing two pure-breeding plants (the

parent generation, P), one purple, one white The offspring (first filial generation, F1), were

referred to as hybrids (mixture of both parents) Monohybrids – hybrid of only one characteristic

All the F1 plants were purple – purple flower color is a dominant trait Thus the white flower color trait is recessive

Mendel’s Experiments

Mendel’s Experimental Design He allowed the F1 to self-fertilize

The resulting second filial generation (F2) showed both dwarf and tall characteristics 705 purple, 224 white, a ratio of 3:1

Mendel did not recognize that the traits were controlled by genes (term coined in 1909)

He did propose that each trait relied on two related, but different, determinants. Alleles represent different forms of a gene

Mendel’s Experiments

Mendel’s Experimental Design Phenotype – observable characteristics

Allele for purple flower plants (P) is dominant over white flower plants (p)

Genotype - combination of alleles an organism possesses Homozygous purple– PP (Both alleles are the same) Heterozygous purple– Pp (Two alleles are different)

Fig. 14-3-3EXPERIMENT

P Generation

(true-breeding parents) Purple

flowers Whiteflowers

F1 Generation

(hybrids) All plants hadpurple flowers

F2 Generation

705 purple-floweredplants

224 white-floweredplants

Fig. 14-5-3

P Generation

Appearance:Genetic makeup:

Gametes:

Purple flowers White flowersPP

P

pp

p

F1 Generation

Gametes:

Genetic makeup:Appearance: Purple flowers

Pp

P p1/21/2

F2 Generation

Sperm

Eggs

P

PPP Pp

p

pPp pp

3 1

Segregation

Law of Segregation During gamete formation the alleles will separate

randomly Fertilization is the fusion of two gametes,

reestablishing the two copies of a gene

Allele for purple flowers

Homologouspair ofchromosomesLocus for flower-color gene

Allele for white flowers

Segregation

Law of segregation explains Mendel’s experiments with four related concepts: 1. Alternate versions of a gene accounts for

variations in inherited characters 2. For each character, an organism inherits

two alleles, one from each parent

Segregation

Law of segregation explains Mendel’s experiments with four related concepts:

3. If two alleles at a locus differ, then one, the dominant allele, determines the organisms appearance, the other, the recessive allele, has no noticeable effect on the organisms appearance.

4. The two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes.

How are each of the 4 parts of Mendel’s law of segregation portrayed in his experiment with crossing peas plants with different flower colors?

• Mendel’s segregation model accounts for the 3:1 ratio he observed in the F2 generation of his numerous crosses

• The possible combinations of sperm and egg can be shown using a Punnett square, a diagram for predicting the results of a genetic cross between individuals of known genetic makeup

• A capital letter represents a dominant allele, and a lowercase letter represents a recessive allele

Segregation

Segregation

Testing the Law of Segregation Along with the 3:1 phenotype ratio, there should also

be a 1:2:1 genotypic ratio 1 PP, 2 Pp, 1 pp

This can be tested by self-fertilizing the F2 to create an F3

The resulting whites should be homozygous The purple F2 plants should be 1/3 homozygous and 2/3

heterozygous The homozygous should produce only purple flower

plants The heterozygous should produce 3 purple flower

plants for each 1 white flower plant (3:1)

Fig. 14-6Phenotype

Purple

Purple3

Purple

Genotype

1 White

Ratio 3:1

(homozygous)

(homozygous)

(heterozygous)

(heterozygous)

PP

Pp

Pp

pp

Ratio 1:2:1

1

1

2

Segregation

Testing the Law of Segregation Another way to test is by using a testcross – cross

any organism with a homozygous recessive If the organism in question is homozygous dominant,

then all progeny will have the dominant phenotype If the organism is heterozygous, then the progeny will be

50% phenotypically dominant, and 50% phenotypically recessive

Practicing a Test Cross

Draw punnett square crossing homozygous white flower pea plant with heterozygous purple flower pea plant. What are the possible offspring? Can you determine the genotypes of the offspring?

Draw punnett square crossing homozygous white flower pea plant with homozygous purple flower pea plant. What are the possible offspring? Can you determine the genotypes of the offspring?

Fig. 14-7

TECHNIQUE

RESULTS

Dominant phenotype, unknown genotype:

PP or Pp?

Predictions

Recessive phenotype, known genotype: pp

If PP If Ppor

Sperm Spermp p p p

P

P

P

p

Eggs Eggs

Pp

Pp Pp

Pp

Pp Pp

pp pp

or

All offspring purple 1/2 offspring purple and1/2 offspring white

Independent Assortment

Mendel analyzed the inheritance of two different traits Homozygous round yellow seeds (YY) crossed with

homozygous wrinkled(rr) green seeds The F1 (dihybrid) were all round yellow seeds

Dihybrid-individuals that are heterozygous for two characters

When the F1 was self-fertilized, the resulting F2 had all four combinations of characteristics

The ratio is very close to 9:3:3:1

Fig. 14-8

EXPERIMENT

RESULTS

P Generation

F1 Generation

Predictions

Gametes

Hypothesis ofdependentassortment

YYRR yyrr

YR yr

YyRr

Hypothesis ofindependentassortment

orPredictedoffspring ofF2 generation

Sperm

Sperm

YR

YR

yr

yr

Yr

YR

yR

Yr

yR

yr

YRYYRR

YYRR YyRr

YyRr

YyRr

YyRr

YyRr

YyRr

YYRr

YYRr

YyRR

YyRR

YYrr Yyrr

Yyrr

yyRR yyRr

yyRr yyrr

yyrr

Phenotypic ratio 3:1

EggsEggs

Phenotypic ratio 9:3:3:1

1/21/2

1/2

1/2

1/4

yr

1/41/4

1/41/4

1/4

1/4

1/4

1/43/4

9/163/16

3/161/16

Phenotypic ratio approximately 9:3:3:1315 108 101 32

Independent Assortment

Punnett squares are used to visualize possible gamete fusions Assumption: Four types of gametes from each dihybrid

parent will be produced in equal numbers

WG Wg wG wg

WG WWGG

WWGg

WwGG

WwGg

Wg WWGg WWgg

WwGg

Wwgg

wG WwGG WwGg

wwGG

wwGg

wg WwGg Wwgg wwGg wwgg

Independent Assortment

Law of Independent Assortment Alleles for one gene can segregate independently of

alleles for other genes Each phenotypic class is made of several different

genotypes Except the homozygous recessive

The genotypic ratio is 1:2:1:2:4:2:1:2:1WG Wg wG wg

WG WWGG

WWGg

WwGG

WwGg

Wg WWGg WWgg

WwGg

Wwgg

wG WwGG WwGg

wwGG

wwGg

wg WwGg Wwgg wwGg wwgg

Independent Assortment

Testing the Law of Independent Assortment This law can be tested by doing a dihybrid testcross

Review: monohybrid heterozygous testcross resulted in a 1:1 phenotypic ratio

Testcross WwGg with wwgg The result is a 1:1:1:1 phenotypic ratio

• Strictly speaking, this law applies only to genes on different, nonhomologous chromosomes

• Genes located near each other on the same chromosome tend to be inherited together

Independent Assortment

Laws of Probability

Mendel’s laws of segregation and independent assortment reflect the rules of probability

Probability

Types of Probability Probability (P) = Number of times an event is

observed (a) / the total number of possible cases (n) P=a/n

Examples: Probability of rolling a 4 on a six-sided dice

P=1/6 Probability of drawing a 7 of clubs from a card deck

P=1/52

Probability

Types of Probability An event that is certain has a probability of 1

P=1/1 An event that is impossible has a probability of 0 An event has a probability of P, the likelihood of the

alternative event is Q=1-P Probability of rolling a 4 on a six-sided dice

P=1/6 Probability of rolling anything else

Q=1-1/6= 5/6 The probability of all possible events must equal 1

P+Q=1

Probability

Types of Probability Mutually exclusive outcomes

Event in which the occurrence of one possibility excludes all other possibilities Rolling a dice, only one side can face up

Independent outcomes Events that do not influence one another

Rolling two dice, the face value of one does not influence the other

Two rules of probability that affect genetics Sum rule and product rule

Probability

Sum Rule When events are mutually exclusive The probability that one of several mutually exclusive

events will occur is the sum of the probabilities Probability of a dice showing either a 4 or 6

P=1/6 + 1/6 = 2/6 = 1/3 The probability increases as the number of possible

outcomes increase Probability of a dice showing any number

P=1/1 Not used for traits expressed on a continuum (human

heights)

Probability

Product Rule When one event is independent of other events The probability that two events will both occur is the

product of their separate probabilities Probability of throwing a die two times and getting a 4 and

then a 6 P=1/6 x 1/6 = 1/32

The probability will decrease as you increase the number of independent events Much like the lottery, the more numbers you need the less

likely you are to win

Probability

Using Probabilities In a monohybrid cross, Dd X Dd (1:2:1), what is the

probability of getting either a homozygous dominant or heterozygous? P= ¼ + ½ = ¾

In a dihybrid cross WwGg X WwGg, what is the probability of getting a round and green seed? P=3/4 x 1/4 = 3/16

Solving Complex Genetics Problems

We can apply the rules of probability To predict the outcome of crosses involving multiple characters

A dihybrid or other multicharacter cross Is equivalent to two or more independent monohybrid crosses occurring simultaneously

In calculating the chances for various genotypes from such crosses Each character first is considered separately and then the individual probabilities are multiplied together

Probability

Branch-Line Approach to Calculate Probabilities Punnett squares require 16 squares for a dihybrid

cross, 64 squares for a trihybrid Punnett squares are useful, but hard to use with more

complex crosses The branch-line approach is based on the law of

independent assortment In branch-line each trait is examined independently

Probability

Branch-Line Approach to Calculate Probabilities AaBb x AaBb To determine the probability of this dihybdrid cross

calculate the probability of each trait and apply the product ruleA phenotype

3/4 (either AA or Aa)

B phenotype ¾ (either BB or Bb)

AB phenotype 9/16 (A-B- genotype)

b phenotype 1/4 (bb genotype)

Ab phenotype 3/16 (A-bb genotype)

a phenotype 1/4 (aa genotype)

B phenotype ¾ (either BB or Bb)

aB phenotype 3/16 (aaB- genotype)

b phenotype 1/4 (bb genotype)

ab phenotype 1/16 (aabb genotype)

Probability

Branch-Line Approach to Calculate Probabilities Use branch-line to determine the probabilities of this

trihybrid cross: AaBbCc x AabbCc

A phenotype 3/4

a phenotype 1/4

B phenotype 1/2

b phenotype 1/2

B phenotype 1/2

b phenotype 1/2

C phenotype 3/4

c phenotype 1/4

C phenotype 3/4

c phenotype 1/4

C phenotype 3/4

c phenotype 1/4

C phenotype 3/4

c phenotype 1/4

ABC phenotype 9/32

ABc phenotype 3/32

AbC phenotype 9/32

Abc phenotype 3/32

aBC phenotype 3/32

aBc phenotype 1/32

abC phenotype 3/32

abc phenotype 1/32

Probability

Branch-Line Approach to Calculate Probabilities The branch-line approach can also be used to determine

genotypes

Statistics

When dealing with probabilistic events, there is a chance the data will cause us to support a bad hypothesis

Mendel’s F2 heterozygous pea plants yielded 788 tall and 277 dwarf. 2.84:1 not exactly 3:1

Is 2.84:1 close enough to represent 3:1?

Statistics

Hypothesis Testing Statistics are used by scientists to summarize data

and test their hypothesis by comparing data predicted results

To determine if the data is consistent with the hypothesis we generate a null hypothesis The null hypothesis assumes the difference between the

observed and expected results are due to chance

Statistics

Chi-Square This test is used when data is distributed among

discrete categories (tall and dwarf plants) Formula for the chi-square

χ is the Greek letter chi, O is the observed number for a category, E is the expected number for the category, and ∑ means to sum the calculations for all categories

χ2=∑(O – E)2

E

Statistics

Chi-Square Example: Mendel observed 787 tall plants and 277

dwarf. Total of 1064. The expected results for a 3:1 ratio are 798 tall and 266 dwarf. Start with figuring the chi-square for tall plants

Then figure it for the dwarf plants

Now sum the two results

(787 – 798)2

798= 0.15

(277 – 266)2

266= 0.45

0.45 + 0.15 = 0.60

Statistics

Chi-Square Example: Use the same data, but assume we were

testing for a 1:1 ratio. How does the chi-square change? Mendel observed 787 tall plants and 277 dwarf. Total of 1064. Start with figuring the chi-square for tall plants

Then figure it for the dwarf plants

Now sum the two results

(787 – 532)2

532= 122.23

(277 – 532)2

532= 122.23

122.23 + 122.23 = 244.45

Statistics

Chi-Square To properly use chi-square values we need to convert

them to a probability value (p) As the number of categories increases so will the chi-

square value (because we sum each category) To help solve this we use degrees of freedom (# of

categories minus 1) Example: two categories would have 1 degree 0f

freedom Consult the chi-square table

Statistics

Chi-Square Table

Probabilities

Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01

1 0.000 0.004 0.064 0.455 1.642 3.841 6.635

2 0.020 0.103 0.446 1.386 3.219 5.991 9.210

3 0.115 0.352 1.005 2.366 4.642 7.815 11.345

4 0.297 0.711 1.649 3.357 5.989 9.488 13.277

5 0.554 1.145 2.343 4.351 7.289 11.070 15.086

6 0.872 1.635 3.070 5.348 8.558 12.592 16.812

Statistics

Chi-Square Table

We will look a p value of 0.05We have only 1 degree of freedom

Probabilities

Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01

1 0.000 0.004 0.064 0.455 1.642 3.841 6.635

2 0.020 0.103 0.446 1.386 3.219 5.991 9.210

3 0.115 0.352 1.005 2.366 4.642 7.815 11.345

4 0.297 0.711 1.649 3.357 5.989 9.488 13.277

5 0.554 1.145 2.343 4.351 7.289 11.070 15.086

6 0.872 1.635 3.070 5.348 8.558 12.592 16.812

Statistics

Chi-Square Table

There is a 0.05 probability of getting a χ2 value of 3.841 or larger by chance alone, given the hypothesis is correct

Probabilities

Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01

1 0.000 0.004 0.064 0.455 1.642 3.841 6.635

2 0.020 0.103 0.446 1.386 3.219 5.991 9.210

3 0.115 0.352 1.005 2.366 4.642 7.815 11.345

4 0.297 0.711 1.649 3.357 5.989 9.488 13.277

5 0.554 1.145 2.343 4.351 7.289 11.070 15.086

6 0.872 1.635 3.070 5.348 8.558 12.592 16.812

Inheritance Patterns

Inheritance patterns are often more complex than predicted by simple Mendelian genetics

The relationship between genotype and phenotype is rarely simple

Extending Mendelian Genetics for a Single Gene

The inheritance of characters by a single gene may deviate from simple Mendelian patterns

The Spectrum of Dominance Complete dominance

Occurs when the phenotypes of the heterozygote and dominant homozygote are identical

Codominance Two dominant alleles affect the phenotype in separate,

distinguishable ways Ex. The human blood group MN is an example of codominance

The Spectrum of Dominance

Incomplete dominance The phenotype of F1

hybrids is somewhere between the phenotypes of the two parental varieties

P Generation

F1 Generation

F2 Generation

RedCRCR

Gametes CR CW

WhiteCWCW

PinkCRCW

Sperm

CR

CR

CR

Cw

CR

CRGametes

1⁄2 1⁄2

1⁄2

1⁄2

1⁄2

Eggs1⁄2

CR CR CR CW

CW CWCR CW

The Relation Between Dominance and Phenotype

Dominant and recessive alleles do not really interact; it is in the pathway from genotype to phenotype that dominance and recessiveness come into play

Lead to synthesis of different proteins that produce a phenotype

Examples: Mendel’s Pea Shape, Tay-Sachs disease

Dominant alleles are not always the most common in a population

• Example: polydactyly (+5 digits)

Multiple Alleles

Most genes exist in populations In more than two allelic

forms

The ABO blood group in humans Is determined by

multiple alleles

Table 14.2

Blood Types

Glycoproteins on surface determine blood type; Important in transfusions/transplants

IA and IB are codominant ii (type O) is recessive to A or BType O = universal donorType AB= universal recipient

Differences in Rh factor (Mom Rh- and baby Rh+) can result in erythroblastosis fetalis

Pleiotropy

Pleiotropy-a gene has multiple phenotypic effects Ex. Hereditary diseases such

as cystic fibrosis & sickle cell disease

Extending Mendelian Genetics for Two or More Genes

Some traits may be determined by two or more genes

epistasis- a gene at one locus alters the phenotypic expression of a gene at a second locus

polygenic inheritance- an additive effect of two or more genes on a single phenotype

EPISTASIS

EX: Coat color in miceB = Black b = brown

C = color deposited in coatc = color NOT deposited

cc-mouse looks white eventhough it has color genes

Image from Biology; Campbell and Reece; Pearson Prentice Hall publishing as Benjamin Cummings © 2005

An example of epistasis

BC bC Bc bc1⁄41⁄41⁄41⁄4

BC

bC

Bc

bc

1⁄4

1⁄4

1⁄4

1⁄4

BBCc BbCc BBcc Bbcc

Bbcc bbccbbCcBbCc

BbCC bbCC BbCc bbCc

BBCC BbCC BBCc BbCc

9⁄163⁄16

4⁄16

BbCc BbCc

Sperm

Eggs EX: Coat color in miceB = Black b = brown

C = color deposited in coatc = color NOT deposited

cc-mouse looks white eventhough it has color genes

Polygenic Inheritance

Many human characters Vary in the population along a continuum and are called

quantitative characters

Quantitative variation usually indicates polygenic inheritance An additive effect of two or more genes on a single

phenotype

POLYGENIC traits are recognizable by their expression as a gradation of small differences (a continuous variation).

The results form a bell shaped curve.

Nature and Nurture: The Environmental Impact on Phenotype

Another departure from simple Mendelian genetics arises When the phenotype for a character depends on environment as well as

on genotype

The norm of reaction Is the phenotypic range of a particular genotype that is influenced by

the environment

Environment influences Phenotype“Nature vs Nurture”

Siamese cats and Himalayan rabbits have dark colored fur on their extremities

Allele that controls pigment production is only able to function at the lower temperatures of those extremities.

Integrating a Mendelian View of Heredity and Variation

Multifactorial characters Are those that are influenced by both genetic and

environmental factors

An organism’s phenotype Includes its physical appearance, internal anatomy,

physiology, and behavior Reflects its overall genotype and unique

environmental historyEven in more complex inheritance patterns

Mendel’s fundamental laws of segregation and independent assortment still apply

Sex Linked Genes

Genes carried on the X chromosome are called X-linked traits.

Red-green colorblindness, hemophilia, an Duchenne muscular dystropy are examples of X-linked traits.

Y-LINKED GENES: Genes carried on the Y chromosome

Y-linked genes only show up in MALES

Hairy pinnaeSRY geneinitiates male sexdetermination

X and y chromosomes

NON-HOMOLOGOUS partners

Pedigrees are diagrams that show how genes are passed on in families over several generations

Pedigrees can be used to predict future offspring in families with genetic disorders

Drawing a pedigree chart

A mutation in an allele that causes a protein to be NON-FUNCTIONAL would appear

recessive to the normal working allele.

Examples of autosomal recessive GENETIC DISORDERS:Phenylketonuria (PKU)Tay-Sachs DiseaseCystic Fibrosis

Phenylketonuria (PKU)

CAUSE: Mutation in gene for an enzyme the breaks down an amino acid called phenylalanine

Build up causes mental retardation

Phenylketonuria (PKU)

ALL babies are tested for PKU before they leave the hospital.

Treatment: Need a diet low in phenylalanine to extend life and prevent mental retardation

CYSTIC FIBROSISCAUSE: Loss of 3 DNA bases in a gene for the ion channel

protein that transports Cl- ions Salt balance is upsetCauses a build up of thick mucous in lungs and

digestive organs, Leads to: Respiratory and digestive complications, increased susceptibility to infections

thick mucous

Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006

http://www.biochem.arizona.edu/classes/bioc460/spring/rlm/RLM36.1.html

Carrier Heterozygous individualThat carries one recessive allele for a genetic disorder

Doesn’t show the disorder themselves,but can pass it on tooffspring

TAY-SACHS DISEASEAutosomal Recessive

CAUSE: Mutation in gene for an enzyme the breaks down a kind of lipid in the developing brain

As these lipids build up in brain infant suffers seizures, blindness, loss of motor & mental function > > > leads to early death.

Found more frequently in people with Jewish, Mediterranean, or Middle Eastern ancestry

Disorders caused by autosomal codominant alleles: SICKLE CELL DISEASE

CAUSE: A changed to T in gene for

Hemoglobin (protein in red blood cells that carries oxygen in blood)

SICKLE CELL DISEASE

SYMPTOMS:Red blood cells become sickle shaped under low oxygen condition in persons with two sickle cell alleles (ss)

Ss=Sickle cell trait Normally healthy, but can suffer some sickle cell episodes

SICKLE CELL DISEASE

Circulatory problemsCells stick in capillariesLoss of blood cells (anemia)Organ damage (brain, heart, spleen)Can lead to DEATH

HUNTINGTON’S DISEASE is AUTOSOMAL DOMINANT

CAUSE:

Extra 40-100 CAG repeats at end of gene on chromosome 4

The more repeats . . . the more severe the symptoms.

HUNTINGTON’S DISEASE

Begins in middle ageCauses progressive loss

of muscle control and mental function

1 in 10,000 people in U.S. have Huntington’s disease

http://www.scielo.br/img/revistas/bjmbr/v39n8/html/6233i01.htm

Huntington’s brain

Normal brain

A person with Huntington’s disease has a _____ chance of passing the disorder on totheir offspring.

Problem:Symptoms of disorder usually don’t show until ____________ . . .

so you don’t know you have it until ________ you have had children.

50%

MIDDLE AGE

AFTER

ACHONDROPLASIA(One kind of Dwarfism)

CAUSE: Autosomal Dominant gene

200,000 “little people” worldwide

One of oldest known disorders – seen in Egyptian art

1 in 25,000 births

DD = lethalDd = dwarf phenotypedd= = normal height

ACHONDROPLASIA(One kind of Dwarfism)

Normal size head and torso; short arms and legs

Problem with way cartilage changes to bone as bones grow

Image from Biology; Campbell and Reece; Pearson Prentice Hall publishing as Benjamin Cummings © 2006