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Page 1: Genetics Chapter 5
Page 2: Genetics Chapter 5

Modifications to Mendelian

Patterns of Inheritance

Chapter 5

Page 3: Genetics Chapter 5
Page 4: Genetics Chapter 5

Variations on Dominance

Mendelian genetics :

• dominant allele requires only 1 copy to determine a

phenotype ; masks expression of recessive allele

• recessive allele requires 2 copies to determine a

phenotype ; expression masked by dominant allele

• crossing 2 monohybrid heterozygotes when

dominance is complete gives a 3:1 phenotypic ratio

sometimes this simple dominant-recessive relationship

does not hold, and a heterozygous genotype does not

always express a dominant phenotype

Page 5: Genetics Chapter 5

Incomplete Dominance in Four-

o’clock Plants (Mirabilis jalapa)

Incomplete dominance : heterozygotes exhibit a

phenotype intermediate to the phenotypes

expressed by the two homozygotes

F2 Phenotypic ratio 1 : 2 : 1

Neither allele is

dominant to the

other : use R1,R2

instead of R, r

Page 6: Genetics Chapter 5

Incomplete Dominance

R1 : specifies red pigment

R2 : specifies no color

• R1, R2 heterozygotes : lighter red because they have only 1 allele that produces color

• Mendelian patterns of inheritance are still followed (1:2:1 ratio in F2), but phenotypes

do not follow patterns of strict dominance

Page 7: Genetics Chapter 5

The ABO Blood Classification

Erythrocytes expressing one (or both) of the ABO

antigens have an antibody against the other antigens

in their serum

Codominance

Codominance : the heterozygote expresses

phenotypes of both homozygotes simultaneously

Page 8: Genetics Chapter 5

The ABO Blood Classification

Codominance

IA IA x IB IB

F1 : 100% IAIB

(type A) (type B)

(type AB)

Phenotype includes both A and B

simultaneously : codominance

F2 : ?

Page 9: Genetics Chapter 5

Pedigree Analysis of the Inheritance

of Tay-Sachs Disease

Analysis of the pedigree indicates that Tay-Sachs is inherited as an autosomal

recessive trait

at the level of viability, shows Mendelian genetics of complete dominance

(only homozygous recessive is not viable)

this disease causes severe neurodegeneration before age 3

homozygous normal and heterozygotes do not have any symptoms

Page 10: Genetics Chapter 5

Hexosaminidase-A in normal and Tay-Sachs

individuals at the level of enzyme activity

The heterozygote individuals possess less activity than the homozygote

dominant individual. The homozygous recessive individual exhibits no activity.

at the level of enzyme activity, shows incomplete dominance

Figure 5.2b

Page 11: Genetics Chapter 5

Hexosaminidase-A in normal and Tay-Sachs

individuals at the level of protein structure

One variant in Tay-Sachs disease produces a protein smaller than the wild-type

protein.

•Afflicted individuals (aa) produce only the smaller protein

•Heterozygote individuals (Aa) produce both proteins

•Homozygous normal (AA) produce the larger protein

at the level of protein structure, shows codominance

Page 12: Genetics Chapter 5

Multiple Alleles

• we have considered genes that have only 2 alleles

• a gene can have more than 2 alleles

- diploid organism can have only 2 alleles, but many

different alleles of any given gene can exist in a

population polymorphic alleles

- in fact, multiple alleles are the rule rather than the

exception

Page 13: Genetics Chapter 5

The ABO Blood Classification

3 alleles : IA, IB, and i

IA, IB : glocosyltransferase enzymes that make A and B

structures on a sugar molecule (H structure) on RBCs

i : nonfunctional enzyme (recessive to IA and IB)

Page 14: Genetics Chapter 5

The ABO Blood Classification

Complete dominance (Mendelian) :

i is recessive to IA and IB

- IA and/or IB will modify the H product and

mask the fact that the i allele is present

Codominance :

IA and IB can be expressed simultaneously

Page 15: Genetics Chapter 5

Fur Color in the Mouse : Controlled by

Several Genes Including Agouti • over 25 different agouti alleles have been identified

• these alleles either exhibit a unique phenotype or have

different genetic interactions with other agouti alleles

Consider 4 of these alleles : A, Aw, at, and a

Heterozygous mice reveal that the Aw allele

is dominant to the other three agouti alleles

(Wild type)

(recessive)

(“white-bellied” allele ; dominant)

(“black&tan” allele)

Page 16: Genetics Chapter 5

Phenotypes of the Different Allelic

Combinations of Mouse Agouti Gene

Wild-type agouti allele (A)

-Dark gray fur color

-Deposition of yellow and black

pigments

White-bellied agouti allele (Aw)

-Dark gray fur on the back, white

or cream on the belly dominant

Nonagouti allele, a, is recessive to

other alleles

Page 17: Genetics Chapter 5

Phenotypes of the Different Allelic

Combinations of Mouse Agouti Gene

These alleles show a hierarchy of dominance :

Aw is dominant to A

Aw and A are dominant to a

. . . and complex patterns of dominance :

A is dominant to at on the back

at is dominant to A on the belly

Page 18: Genetics Chapter 5

Testing Allelism

* There may be many different variations of the

2 alleles that code for a gene (such as with

the gene for mouse fur color)

* Also, mutations in different genes may cause

similar or identical phenotypes

To study how phenotypes are inherited, need to

know which mutations are different

versions of the same gene (alleles) and which

are in completely different genes

Page 19: Genetics Chapter 5

Partial map of Drosophila chromosomes

ruby and scarlet are

2 different mutations

of 2 different genes on

different chromosomes,

but they both produce

a similar eye color

Page 20: Genetics Chapter 5

Complementation test in flies: determine

whether 2 recessive mutations are alleles

•Cross flies with homozygous recessive mutations, examine progeny :

1) wild-type phenotype : mutations complement each other

mutations are in different genes

arrangement of the two recessive mutations on different

chromosomes is called the trans configuration (each homolog contains 1

mutation)

wingless wingless

mutant gene 1, wt gene 2

wt gene 1, mutant gene 2

wild-type

: mutation

Page 21: Genetics Chapter 5

Noncomplementation (1)

2) none of the F1 progeny exhibit the wild-type phenotype:

mutations do not complement each other

mutations are in different alleles of the same gene

wingless wingless

mutant gene 1, wt gene 2

mutant gene 1, wt gene 2

wingless

Page 22: Genetics Chapter 5

Noncomplementation (2)

• Cross flies with nonidentical recessive mutations, examine

progeny

- F1 progeny : small, crinkled wings (not wild-type trait)

these mutations do not complement each other and must be

alleles

wingless Small, wrinkled wings

Very Small, wrinkled wings

mutant gene 1, wt gene 2

mutant gene 1, wt gene 2

Page 23: Genetics Chapter 5

(cis configuration is a control for test

to make sure a wt can be produced)

Parents : i) both mutants on same chromosome i) 1 mutant on 1 chromosome

ii) wild-type ii) other mutant on other chromosome

The Cis-Trans Test • examine a large # of mutants and place them in complementation groups

** identifies : 1) mutations in separate genes and 2) independent alleles of a

single gene

(trans configuration determines if

mutations are allelic or not)

Page 24: Genetics Chapter 5

The Cis-Trans Test

can use this test to determine how many genes control

a biochemical pathway

- try to mutate every gene in a pathway, then use cis-trans test

to determine how many distinct genes code for proteins that

affect the same mutant phenotype

helps to treat genetic diseases

Page 25: Genetics Chapter 5

Lethal Alleles

• cause a deviation from expected 3:1 phenoptypic

ratio in a monohybrid cross

• caused by certain genotypes that result in death

and are not seen in the progeny

• these alleles can still follow standard Mendelian

dominant-recessive relationships but will skew

Mendelian phenotypic ratios

Page 26: Genetics Chapter 5

Relationship of yellow (AY) allele to the

wild-type agouti (A) allele

• Homozygous agouti X yellow : always produces ½ yellow and

½ agouti mice (expected 1:1 ratio for a monohybrid test cross)

because the agouti mice are known to be homozygous,

yellow mice must be heterozygous with yellow dominant to

agouti

(1)

2 important crosses :

? we already know agouti

mice are genotypically

AA

Page 27: Genetics Chapter 5

•Yellow X yellow 2/3 yellow, 1/3 agouti

•Expected results phenotypic ratio 3:1 yellow to agouti

(monohybrid cross of heterozygotes)

•Possible explanations?

Relationship of yellow (AY) allele to the

wild-type agouti (A) allele

(2)

Page 28: Genetics Chapter 5

Examination of pregnant mice revealed 25% of the

developing embryos were dead

•Homozygous yellow allele is lethal: AyAy

•2/3 of the surviving mice will be yellow (heterozygous)

•1/3 will be agouti (homozygous)

Relationship of yellow (AY) allele to the

wild-type agouti (A) allele

Page 29: Genetics Chapter 5

AY allele is dominant for coat color, recessive for lethality

Pleitropy : when a single mutation causes multiple

phenotypes

Relationship of yellow (AY) allele to the

wild-type agouti (A) allele

Page 30: Genetics Chapter 5

Other Lethal Mutations • Dominant lethal mutations

Can only exist if :

i. Lethality is only expressed after sexual maturity

- example : Huntington disease (onset : after age 40)

ii. Incomplete penetrance : a phenotype is not

expressed for a certain genotype

• Deleterious mutations

- reduce viability without always causing death ;

depends on environment

- example : temperature sensitive mutations in flies

Page 31: Genetics Chapter 5

What about when a specific genotype

does not express the expected

phenotype ?

3 possibilities :

i. incomplete penetrance

ii. variable expressivity

iii. 1 gene affects the expression of

another gene

Page 32: Genetics Chapter 5

Incomplete Penetrance: A Percentage of

individuals with a genotype do not express

the expected phenotype

Only 60% of the individuals are expressing the expected

phenotype ; it is 60% penetrant

• Most genotypes are 100% penetrant

(all are homozygous recessive genotype)

Page 33: Genetics Chapter 5

Variable Expressivity : the range of

phenotypes associated with a specific

genotype is increased

(all are homozygous recessive genotype)

All express a mutant phenotype, but there is a range of

these phenotypes (light red to complete white)

- these phenotypes are associated with a single

mutant allele (caused by different expression levels?)

Page 34: Genetics Chapter 5

Incomplete Penetrance and Variable

Expressivity : Polydactyly

•Autosomal dominant trait

•Extreme manifestations characterized

by an extra digit on each hand and one

or two extra toes on each foot

•Variable expressivity:

•Extra toes or extra fingers

•Portion of an extra digit – thumb

•Partial extra little toe

incomplete penetrance and variable expressivity : affected by genetics and

environmental factors

Page 35: Genetics Chapter 5

Genotypic Interactions

Mendelian Inheritance :

Cross between 2 independent traits should

always produce 9:3:3:1 phenotypic ratio

- 9/16 have both dominant phenotypes

- 1/16 has both recessive phenotypes

Variations in phenotypes or in ratios

indicates that 2 genes may be interacting

Page 36: Genetics Chapter 5

•Cross Rose-combed chicken X pea-

combed chicken (or vice versa)

•F1 offspring are walnut-combed

•Cross F1 Heterozygous walnut-

combed chickens

•F2 progeny – 9:3:3:1

indicates 2 genes are involved

•9/16 walnut-combed fowl

•3/16 rose-combed fowl

•3/16 pea-combed fowl

•1/16 single-combed fowl

prediction of genotypes ?

2 Genes Contribute to a Single Phenotype: 1) mixed phenotypes : genetic interactions in the

combs of chickens

Page 37: Genetics Chapter 5

Interpretation:

•Dominant alleles for each gene: (R-)

(P-) : walnut-combed

•Dominant rose allele (R-) and

homozygous pea (pp) – rose-combed

•Dominant pea allele (P-) and

homozygous rose (rr) – pea-combed

•Homozygous recessive (pprr) for

both alleles – single-combed fowl

2 genes affect 1 phenotype

2 Genes Contribute to a Single Phenotype: 1) mixed phenotypes : genetic interactions in the

combs of chickens

Page 38: Genetics Chapter 5

•Crossing 2 white corn varieties yields all

purple kernels in F1

•F1 cross yields 9/16 purple

7/16 white ???

-Must be dealing with 2 traits, each with 2

alleles because ratios are in sixteenths

-9:7 is variation of 9:3:3:1 ratio

-genotypic possibilities in Punnet square :

purple color appears when at least one

dominant allele for both genes is present

white color appears when one or both

genes only have recessive alleles

2 Genes Contribute to a Single Phenotype: 2) complementary gene action : genetic interactions in

the color of corn kernels

Page 39: Genetics Chapter 5

2 Genes Contribute to a Single Phenotype: 2) complementary gene action : genetic interactions in

the color of corn kernels

• a dominant allele for 2 different genes must be present to

produce a phenotype (purple)

• without this allele at either or both genes, white color results

can order these genes in a biochemical pathway

Colorless precursor colorless intermediate purple product

Gene A

catalyzes

Gene B

catalyzes

Page 40: Genetics Chapter 5

Shepherd’s purse plant (Capsella bursa-pastoris)

Indicates heart-shaped is dominant to narrow;

should be a simple monohybrid cross

???

- expect 3:1, 1:2:1, or 2:1 for monohybrid

F2 ratio

Ratio is modification of 9:3:3:1 2 genes are involved

2 Genes Contribute to a Single Phenotype: 3) duplicate gene action : 2 genes produce a

phenotype ; both genes appear equivalent

Page 41: Genetics Chapter 5

15:1 ratio indicates that the dominant allele (for either the A or B gene) is

sufficient to produce the heart-shaped phenotype.

•the enzymes encoded by the A and B alleles are equivalent in the biochemical

pathway :

• 2 different genes encode identical enzymes

one of these enzymes can have a slightly different function ;

creates gene families with similar function (as with -globin family

composed of 5 genes, each expressed at different times in development)

2 Genes Contribute to a Single Phenotype: 3) duplicate gene action : 2 genes produce a

phenotype ; both genes appear equivalent

Page 42: Genetics Chapter 5

•Crossing a pure-breeding black

mouse with a pure-breeding

albino mouse produce 100%

agouti mice

•The albino phenotype results

from the absence of pigment in

the hair

2 Genes Contribute to a Single Phenotype: 4) epistasis : phenotype produced by 1 gene (epistatic

gene) masks the phenotype produced by 2nd gene

(hypostatic gene)

Coat Color in Mice: Black, Albino, or Agouti

* * Indicated by modified 9:3:3:1 ratio with only 3 phenotypes

Page 43: Genetics Chapter 5

F2 9:3:4 ratio is a modification of 9:3:3:1 recessive epistasis

F2 progeny:

•9/16 agouti A-C-

•3/16 black aaC-

•4/16 albino

(3/16 A-cc and 1/16 aacc)

2 Genes Contribute to a Single Phenotype: 4) epistasis : phenotype produced by 1 gene

masks the phenotype produced by 2nd gene

any genotype with cc is albino and masks phenotype of the A

gene (agouti or black) ; without this, the A gene expresses (A>a)

Page 44: Genetics Chapter 5

Color of Summer Squash Can Be White, Yellow, or Green

P1: pure-breeding white X

pure-breeding green

•F1 progeny : all white

squash

•F1 is self-crossed

2 Genes Contribute to a Single Phenotype: 4) epistasis : phenotype produced by 1 gene

masks the phenotype produced by 2nd gene

Page 45: Genetics Chapter 5

2 Genes Contribute to a Single Phenotype: 4) epistasis : phenotype produced by 1 gene

masks the phenotype produced by 2nd gene

F2 12:3:1 ratio is a modification of 9:3:3:1 dominant epistasis

dominant A allele masks phenotype of B gene

Page 46: Genetics Chapter 5

Mechanism of Epistasis

Page 47: Genetics Chapter 5

Defects in the Biochemical Pathway Producing

Melanin Results in Albino Mice

Mouse coat color :

Both the agouti

and black

phenotypes need

melanin in order to

be expressed

Albinism is caused by defects pathway

for melanin production :

Page 48: Genetics Chapter 5

The Epistatic Gene Functions Early in the Pathway

Resulting in the Albino Phenotype for cc Mice

The recessive c allele is responsible for the albino pigment by blocking

the pathway ; its product is required by next gene (A) in making the

yellow coat color

Page 49: Genetics Chapter 5

In biochemical pathways involving epistasis :

epistatic gene functions earlier than hypostatic gene

genetic analysis complements a biochemical

analysis

Page 50: Genetics Chapter 5

Suppression of the Vestigial

(vg) Wing Phenotype in

Drosophila :

In this case the su(vg)

recessive mutation “pushes”

the vestigial phenotype

toward the wild-type

recessive suppressor

2 Genes Contribute to a Single Phenotype: 5) suppression : one gene “pushes” the mutant

phenotype of a 2nd gene towards wild-type

phenotype

classes:

+ +

+ - -

- - - -

- - + -

Page 51: Genetics Chapter 5

Suppressors :

1) May have no other phenotype than to suppress

another gene’s mutant phenotype

dihybrids produce only 2 phenotypes

2) Only make the second gene’s mutant phenotype

more like wild-type (no effect on wild-type allele)

3) Can be either dominant or recessive, may suppress a

dominant or recessive allele

4) May have both dominant or recessive phenotypes

Page 52: Genetics Chapter 5

A suppressor gene may

encode a protein that

specifically interacts with

another protein :

Two genes (m and su) encode

different proteins that interact

to convert a substrate to a

product and produce the wild-

type phenotype

* finding these suppressor mutations

is a powerful way to identify which

gene product forms a protein

complex with another gene product

Possible Mechanisms of Suppression:

Page 53: Genetics Chapter 5

Other mechanisms (nonspecific; can act

on several other genes):

1) increase or decrease expression

level of gene

2) affect translation of an mRNA

- prevent a stop in translation

induced by nonsense mutation

Possible Mechanisms of Suppression:

Page 54: Genetics Chapter 5

Modified 9:3:3:1 ratio in F2 generation

2 genes are interacting in the

phenotype

epistasis : often only 3 phenotypes

9:3:4

12:3:1

suppression : often only 2 phenotypes

13:3

Page 55: Genetics Chapter 5

Epigenetics

heritable modification of gene

function without a change in

DNA sequence

Page 56: Genetics Chapter 5

Example of epigenetics :

mosaicism in human females

If the female is heterozygous for a sex-linked gene, she is mosaic

(composed of two or more genetically distinct tissues or cell types)

- some tissues express dominant allele, other tissues express recessive

allele

Random inactivation of X chromosome during embryonic

development in female mammals -some cells (and developing tissues) have paternal X chromosome

inactivated, others have maternal X chromosome inactivated

Page 57: Genetics Chapter 5

Epigenetics

Barr bodies

http://www.mun.ca/biology/scarr/Barr_Bodies.html

Page 58: Genetics Chapter 5

histone

modifications

DNA methylation

adapted from http://www.landesbioscience.com/curie/chapter/4639/

Epigenetics

Effects of diet and behavior on epigenetics and gene expression :

http://video.pbs.org/video/1525107473/

methyl group