Extensions of Mendelian Genetics_2014

7/24/2019 Extensions of Mendelian Genetics_2014

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Extensions of Mendelian Genetics

Dr. E. Hannig

BIOL3301_Sept. 2014

Ratio Review

3:1 Phenotypic Ratio, Monohybrid cross

WWx ww P

Ww F1(100% express dominant phenotype)

self

WW

Ww F2(3:1 phenotypic ratio)Ww

ww

F2: 2 phenotypes, 3 genotypeso Self the F2round seeds.1/3 true-breeding, 2/3 yield round and wrinkled in

3:1 ratio

o Self F2wrinkle seeds.all are true-breeding for the recessive phenotype

Therefore, the 3:1 phenotypic ratio masks an underlying 1:2:1 genotypic ratio, where

the 2/4 represents the heterozygous genotype

That said,for some traits, the F2may show a 1:2:1 phenotypic ratio if the heterozygotesshow a different phenotypefrom the homozygous dominant and homozygous recessive

phenotypes.

In this case, theRalleleis said to be incompletely

dominant.


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9:3:3:1 ratio is characteristic of a dihybrid cross involving two independently-assorting

genes that determine discernable traits (e.g., seed color and seed shape, etc.)

RRYYx rryy P [Rround and Yyellow are dominant]

RrYy F1

9/16 round, yellow

3/16 round, green

3/16 wrinkled, yellow F2

1/16 wrinkled, green

The F2 generation (from selfing the F1s):

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Gene Interaction Leads to Modified Dihybrid Ratios

What if, in a dihybrid cross, the two genes interact with each other in determining a

single phenotype (e.g., flower color or seed shape)??http://www.nature.com/scitable/topicpage/epistasis-gene-interaction-and-phenotype-effects-460

Examples of Dihybrid (Digenic) RatiosRatio Description Name(s) of Relationship

(Used by Some Authors)

9:3:3:1 Complete dominance at both gene pairs; new

phenotypes result from interaction betweendominant alleles, as well as from interaction

between both homozygous recessives

Not named because the ratio

looks like independentassortment

9:4:3 Complete dominance at both gene pairs;

however, when one gene is homozygous

recessive, it hides the phenotype of the other

gene

Recessive epistasis

9:7 Complete dominance at both gene pairs;however, when either gene is homozygous

recessive, it hides the effect of the other gene

Duplicate recessive epistasis


however, when one gene is dominant, it

hides the phenotype of the other gene

Dominant epistasis

15:1 Complete dominance at both gene pairs;

however, when either gene is dominant, ithides the effects of the other gene

Duplicate dominant epistasis

13:3 Complete dominance at both gene pairs;however, when either gene is dominant, it

hides the effects of the other gene

Dominant and recessiveepistasis


however, when either gene is dominant, it

hides the effects of the other gene

Duplicate interaction

7:6:3 Complete dominance at one gene pair and

partial dominance at the other; whenhomozygous recessive, the first gene is

epistatic to the second gene

No name

3:6:3:4 Complete dominance at one gene pair and

partial dominance at the other; when

homozygous recessive, either gene hides theeffects of the other gene; when both genes

are homozygous recessive, the second gene

hides the effects of the first

No name

11:5 Complete dominance for both gene pairs

only if both kinds of dominant alleles arepresent; otherwise, the recessive phenotype

appears

No name

3
http://www.nature.com/scitable/topicpage/epistasis-gene-interaction-and-phenotype-effects-460http://www.nature.com/scitable/topicpage/epistasis-gene-interaction-and-phenotype-effects-460http://www.nature.com/scitable/topicpage/epistasis-gene-interaction-and-phenotype-effects-460


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Interacting Genes in the Same Pathway

Experiment:

Isolate mutants (screen/selection)

Is the mutation dominant or recessive? [here, we want recessive]

Does the mutant strain contain only one mutation relevant to the phenotypescreened/selected for? [here, the answer must be yes]

Construct strains true-breeding for the phenotype

Cross (mate) true-breeding mutants with each other [e.g., true-breeding mutant #1 x true-

breeding mutant #2]

What is the phenotype of the progeny??

This is the standard way of doing complementation testing to determine if two mutants

harbor mutations in the same or in different genes.

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Example I. Complementation(Complementary Gene Action)

o Complementary Gene Action yields a 9:7 modified dihybrid ratioo If two true-breeding white-flowered plants (the mutant, recessive phenotype),

when crossed with each other, yield blue flowered plants (the wild-type, dominant

phenotype) in the F1, the mutations in each of the plant are said to complement

each otherthe implication here is that the mutations in the two white-flowered plants lie in different genes.

o OTOH, if the cross of two true-breeding white-flowered plants yields white-flowered F1s, the mutationsfail to complementeach other..the implication is

that the mutations in the two white-flowered plants lie within the same gene.

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Lets call the w1 gene from above gene C, and the w2gene above well call gene Pin the

example below. They are different genes.

Thus, a CCpp(white) x ccPP(white) cross will yield CcPpdihybrid F1progeny.

Because the mutants complement each other, they exhibit complementary gene action indetermining flower color.

That is, :

Now lets self the F1blue-flowered (due to anthocyanin production) plants.

In this case, the F1 genotype is CcPp, so the cross isCcPpx CcPp[you will recognizethis as a dihybrid cross; unlike the prior dihybrid crosses we discussed, this one

involves two genes that determine the SAMEphenotype]

Analysis of the F2:

Genotype Flower Color Enzyme Activities

9 C_P_ Flowers colored;anthocyanin produced

Functional enzymes from both genes

3 C_pp Flowers white;

no anthocyanin produced

p enzyme non-functional

3 ccP_ Flowers white;no anthocyanin produced

c enzyme non-functional

1 ccpp Flowers white;no anthocyanin produced

c and p enzymes non-functional

Phenotypically, this gives a 9:7 ratio (pigment : no pigment)this is an example of amodified dihybrid ratio [modified from 9:3:3:1]

Female Gametes

CP Cp cP cp

Male

Gametes

CP CCPP CCPp CcPP CcPp

Cp CCPp CCpp CcPp Ccpp

cP CcPP CcPp ccPP ccPp

cp CcPp Ccpp ccPp ccpp

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Modifications of the 9:3:3:1 dihybrid ratio is indicative of two genes

involved in determining a single phenotype!

Lets consider other examples..

Duplicate Geneso Instead of two genes

acting sequentially in a

pathway, what if two

genes act at the same stepin a pathway such thateither alone is sufficient

to complete the pathway?o Kernel color in wheat is a

good example [purple vs

non-purple]

here, either geneAorgeneBis sufficient

The cross: Purple x non-purple, both true-breeding

(P) F1 F2AABB(purple) x aabb(not purple) AaBb(purple) self ?????

Female Gametes

B b aB ab

Male

Gametes

B AABB AABb AaBB AaBb

b AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

9/16A_B_ Purple kernels, both A and B functional

3/16A_bb Purple kernels, A functional, B is not; either A or B alone is sufficient3/16 aaB_ Purple kernels, A non-functional, B functional; either A or B is sufficient.

1/16 aabb Non-purple, neither A nor B functional.

The modified dihybrid ratio in the case of duplicate genes (functionally, if notstructurally) is 15:1.

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Recessive epistasis: Generation of an F2 9:4:3 dihybrid ratio.

Epistasis (definition)

o Literally, stand upon;o In the original sense, defined by Bateson, epistasis arises when the effects of

alleles at one locus are blocked by the presence of a specific allele at anotherlocus

o the ability of one mutation to override another in a double mutant..here, we areusually dealing with 2 different mutant phenotypes in addition to wild type.

Mouse coat color primer

o GeneAand gene Care two unlinked genes involved in mouse coat color [otherloci also affect coat color].

o Two true-breeding colors with respect to theAand Cgenes: agouti (AACC) andwhite (aacc)

o The agouti (normal) color begins as black pigmentation of coat hairs, followed

by yellow banding of the hair shafts.

Cross of true-breeding agouti and white mice:Crossing true-

breeding agouti and

white strains yieldsall agouti F1

progeny.

Intercrossing F1males and females

yields yet anothermodified dihybridratio, 9:3:4,

indicative of the

presence of 2distinct genes

involved in the coat

color phenotype

Note the appearance

of a third phenotype

(black coat color)..specifically in a 3/16 ratioindicating that these mice areeitherA-/ccor aa/C-.

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The 4/16 white progeny indicate the genotype of of those mice (1/16 of

the total F2) isaacc.

9/16A-C-[agouti]3/16A-cc[albino]

3/16 aaC-[black]1/16 aacc[albino]

o The presence of the ccrecessive genotype therefore blocks theexpression of the black coat color phenotype [aa].

o In genetic terms, ccis epistatic to aa, since the double mutantexpresses the cphenotype

o we can interpret this as the recessive c phenotype standing upon the

aa phenotype, and the C gene acting before the A gene in the pathwayleading to the agouti coat color as follows:

C A

albino black agouti

Here, thecgene is referred to as the epistatic gene; theagene is thehypostatic gene in thec/ainteraction.

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Heres another gene involved in coat color, theEgene. eemice show arecessive yellow phenotype. An allele of theAgene,A

y, yields a dominant

yellow coat phenotype. The epistasis is complicated because theAgene andtheEgene act in an antagonistic manner (see below)

A agouti-signaling protein

C tyrosinase

E melanocortin-1 receptor (Mcr1)

The biochemistry of this pathway within the melanocyte has been elucidated. Activation

of MC1R (theEgene) turns on production of eumelanin (via the Cgene product), as

opposed to the default production of pheomelanin. The agouti protein acts an antagonistto MC1R, leading toperiodicactivation of the agouti protein (theAgene) and banding of

color on individual hairs. Disruption of other loci, such as the classic tyrosinase locus

(Tyr; previously known as albino..the Cgene), destroys function of the entire pathway.The specific interactions between the proteins in this pathway are representative of

functional epistasis.

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Other coat color genes in mice:http://www.ncbi.nlm.nih.gov/books/NBK21249/

The B geneThis gene determines the color ofpigment. There are two major alleles,

Bcoding for black pigment and bforbrown. The alleleBgives the normal

agouti color in combination withAbut gives solid black with a/a. The

genotypeA/ b/bgives a streaked

brown color called cinnamon,anda/ab/bgives solid brown. In horses,

the breeding of domestic lines seems

to have eliminated theAallele that determines the agouti phenotype, although certainwild relatives of the horse do have this allele. The color we have called brownin mice is

called chestnutin horses, and this phenotype also is recessive to black.

The D geneTheDgene controls the intensity of pigment specified by the other coat color genes. The

genotypesD/DandD/dpermit full expression of color in mice, but d/ddilutes thecolor, making it look milky. The effect is due to an uneven distribution of pigment in the

hair shaft. Dilute agouti, dilute cinnamon, dilute brown, and dilute black coats all are

possible. A gene with such an effect is called a modifier gene. In horses, theDallele

shows incomplete dominance. Cases of dilution in the coats of house cats also arecommonly seen.

The S geneThe Sgene controls the presence or absence of spots by controlling the migration of

clumps of melanocytes (pigment-producing cells) across the surface of the developingembryo. The genotype S/ results in no spots, and s/sproduces a spotting pattern called

piebaldin both mice and horses. This pattern can be superimposed on any of the coat

colors discussed earlier, with the exception of albino, of course. Piebald mutations arealso known in humans.

Thus, normal coat appearance in wild miceis produced by a complex set of interacting

genes determining pigment type, pigment

distribution in the individual hairs, pigment

distribution on the animals body, and thepresence or absence of pigment. Such

interactions are deduced from modified

ratios in dihybrid crosses. The figureillustrates some of the pigment patterns in

mice. Interacting genes such as these

determine most characters in any organism.

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http://www.ncbi.nlm.nih.gov/books/NBK21249/http://www.ncbi.nlm.nih.gov/books/NBK21249/http://www.ncbi.nlm.nih.gov/books/NBK21249/


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Coat color in Labrador retrievers..yet another example of recessive epistasis

Bis the dominantBlack allele.

bthe recessive brown allele.Eis the dominant Extension gene, in conjunction withB/bcauses a dog to be black or

brown. eeresults in cream color.

Suppression: Dominant

Example: Malvidin (an anthocyanin) production inPrimula

Requires the product of the Kgene..but, the action of Kcan be suppressed (negated, turned off, masked) by the presence of a non-allelic

suppressor,D, whos effect is dominant [i.e., K-D-strains fail to produce malvidin.

P KKdd(malvidin) x kkDD (no malvidin)

F1 KkDd(no malvidin) !!

F2 9/16 K-D- (no malvidin)

3/16 K-dd (malvidin)3/16 kkD- (no malvidin)1/16 kkdd (no malvidin)

Thus, dominant suppression of theKgene by theDsuppressor yields a 13:3

(mutant : wild-type) dihybrid ratio.

Ques: If the action of the suppressor was recessivewhat would be the phenotypes

of the F1plants? the modified dihybrid ratio???

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Once you have answered the last question, you should realize that, while

modified dihybrid ratios indicate the involvement of two genes in

determining the same phenotype, they are more likely to rule out

certain types of gene interactions than they are at identifying a single,

specific mechanism.

Suppression: recessiveo A suppressor is an allele that reverses the effect of a mutation of

another gene, resulting in the normal (wild-type) phenotype.

Example: purple eye color inDrosophila

Experiment:

o Start with wild-typeDrosophila(red eyes) and isolate (screen for) rare mutantsthat show a purple eye phenotype.in this case, due to a single recessive

mutation in thepdgene.

o Construct true-breeding strains for the Pd-phenotype (pd/pd)o Reversion analysis (whats that??) to isolate wild-type strains from the true-

breedingpd/pdstrains

o if reversion is due to an unlinked suppressor gene (su) that acts in arecessive manner (therefore, the genotype would be pd/pd; su/su),then

Note: alternatively, I could

cross the suppressor strain

(pd/pd su/su) with a wild-

type strain (pd+/pd+;su+/su+)

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How do suppressors work?Some possible explanations:

1.Nonsense suppressionthe pd allele contains a nonsense mutationin the protein-coding sequence; su encodes a tRNA suppressor gene

that decodes (at a low rate) nonsense codons

2. The Pd protein normally forms an active complex with the Su protein;however, the pd mutation encodes an altered Pd protein such that it no

longer interacts with Su..so no active complex is formed. The sumutation alters the Su protein such that is restores the interaction Su

with the mutant Pd protein. alone!

Notice that, in this mechanism: only certain

mutations in su will suppress the pd mutation,

and the su mutation alone may have the samephenotype as the pd mutation

3. Two pathways lead to the production of the correct eye colorpigmentone is constitutive, one regulated. The pd mutation knocks

out flow through the constitutive pathway; the su mutation, however,alters the regulated pathway such that it now becomes constitutive.

4. A variation on mechanism #3 is that the su mutations bypasses thepathway blocked by the pd mutation by channeling intermediates

beyond the block from related pathways.5. Still another mechanism is that the pd mutation reduces the function

of the Pd protein to ~25%, which is insufficient to give red eyes. The

su mutation increases the expression (transcription) from the pd genesuch that, even though pd produces protein with 25% function, thereis now more of it.sufficient to give a red-eye phenotype

6. ..I could go on, but I think you get the point

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Interacting Genes in Different Pathwayso Here we have a situation where two genes are necessary for determining color;

however, in this case, the pathways are separate and both end products are

necessary for normal or wild-type color.

Color in corn snakes: normal (camouflage), black, orange, and albino are

true-breeding phenotypes.

(P) true-breeding orange x true-breeding black(F1) all normal

..intercross F1 males and females

Top left: Normal (aka

camouflaged)

Top right: Anerythstic (unable to

make red/orange pigment)

Bottom left: Amelanistic (unable

to make black pigment)

Bottom right: Albino (unable to

make either pigment)

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Lethality and Pleiotrophy

Pleiotrophy = single allele having more than one phenotypic effect

Ayallele dominant for yellow coat color

recessive lethal

QUES: How would I determine that all of my yellow mice areAy/A

heterozygotes?

o In addition to a dominant yellow coat color, other phenotypes associated withA

y/Amice include obesity, insulin-resistant type II diabetes, and increased

propensity to develop a variety of spontaneous and induced tumors.

o Molecularly, theAyallele is the result of a deletion near theAlocus that results inoverexpression of the agoutigene. Lethality is likely due to the deletion extending

into neighboring gene(s) and not to overexpression of agoutiitself.

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The Manx cat

The tailless Manx is the result of a

genetic mutation that was thenintensified by the cats remote

location on the Isle of Man, off the

coast of Britaino The Manx gene is a dominant

allele of the Mgene (ML) that

interferes with spinal

development and shortens thespine of the cat, however not

always to the same extent - creating various types of Manx

Synthetic Lethality

The search for synthetic lethals can be used to search for additional genes that affect thepathway/mechanism you are studying. OTOH, it often yields mutations that alter a

(completely) different pathwaymutant 1 makes cells sick for one reason, mutation 2

also makes the cells sick, but for a different reason. Combining sick mutation 1 with sick

mutation 2 may make cells too weak to survive.

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Patterns of Dominance

Complete dominance

This is what we have been considering to date [example:RR (red) x rr(white) All redF1s]

Possible explanations:

1. AlleleRproduces 20 units of an enzyme..lets say its required to make the amino

acid leucine

Recessive allele rproduces 12 units of the enzymeDominant alleleR

dproduces 0 units of the enzyme

.but to give a wild-type Leu+phenotype, an organism must produce 30 units of an

enzyme; anything less, the organism is Leu-and requires you add leucine to the media to

grow.

RR= 40 units of enzyme; Leu+

rr= 24 units of the enzyme; Leu

-

Rr= 32 units of the enzyme; Leu+

RRd= 20 units of the ezyme; Leu

-

2. Consider transcriptional activator proteins. These typically contains 2 protein domains

within the same polypeptideone that binds DNA, the other that attracts RNA

polymerase to the gene so it can be transcribed.- a deletion allele of the gene might be recessive, as only one copy of the gene is

necessary.

- how about a missense mutation in the gene that makes a protein that can bind DNA, butlacks the transcriptional activation domain; this type of allele might be dominant, as it

can bind to DNA perfectly well and, in doing so, block access of the wild-type protein (in

a heterozygous strain) to the DNA. Another possible scenario is the poisoned complexsituation, where the active protein is a complex of multiple polypeptides (either the same

or different ones)..here, incorporation of a mutant/altered protein might prevent the

complex from functioning normally, potentially yielding an inactive complex.

3..and there a quite a few more possibilities

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Incomplete dominance:o heterozygote phenotype is intermediate between a homozygous

dominant and homozygous recessive phenotype

Incomplete dominance can run the gamut between the rrand theRR

phenotypes, as shown below:

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Think of incomplete (partial) dominance as a scale. If A1and A2are two

alleles of the same gene:o

If A1is completely dominant, then an A1A2heterozygote has thesame phenotype as an A1A1homozygoteo If A2is completely dominant, then an A1A2heterozygote has the

same phenotype as an A2A2homozygoteo An A1A2phenotype that is roughly in the middle is sometimes

referred to as no dominance; however, for our purposes, unless youhave quantitative data that the phenotype is at the halfway pointbetween A1and A2, well simply call this situation..and anyphenotype between A1A1and A2A2 homozygotes partial

(incomplete) dominance

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Co-dominance

Blood types are an excellent example of co-

dominance, where in the heterozygoteexpresses both phenotypes (i.e., not some

intermediate phenotype)The three common major blood group

alleles areIA,I

B, and i

Possible combinations of major blood group alleles and their phenotypes:

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Bombay phenotype: the H gene

Now lets look at a cross in which we consider both gene sets together:

The cross:IAI

BHh x I

AI

BHh [consider that these gene pairs assort

independently from each other] :

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Ratios look familiar? Maybe not.we havent seen a 3:6:3:4 ratio before,

but this is an excellent example of epistasis. Hmmm.I thought that was

9:3:4..well, it was in the example that we considered (mouse coat colorand theAand Cgenes). The difference lies in the fact that, in the present

situation, we are dealing with co-dominant alleles (IAand I

B).whereas in

the case of theAand Cgenes..we were not!

In the above example:

1. Which is the epistatic allele?2. Which is the hypostatic allele?3. What is the predicted order of execution, assuming both genes operate

in the same linear pathway that leads to major blood group type?

Biochemical basis of

ABO blood group

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The sickle cell phenotype

The mutation (E6V) is a change in the -globin gene that changes aminoacid 6 from glutamic acid (an acidic amino acid) to valine (a hydrophobicamino acid). The charge loss and its replacement by a hydrophobic amino

acid lead to a decrease in the solubility of hemoglobin and the resultingsymptoms.

Phenotype: anemia [HbAdominant]

HbAHb

A not anemic

HbSHb

S not anemic

HbAHb

S anemic

Phenotype: RBC shape

[HbAincompletely

dominant]

HbAHb

A normal

HbSHb

S RBCs sickle

HbAHb

S RBCs may sickle

under low pO2 (high

altitudes), dehydration, orincreased atmospheric

pressure (e.g., scuba diving),but not all individuals with

SCT (sickle cell trait) areaffected to the same degree.

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Phenotype: production of specific hemoglobin (biochemical phenotype)

[HbAand Hb

Sare co-dominant]

HbAHbA Hemoglobin A onlyHb

SHb

S Hemoglobin S only

HbAHb

S Hemoglobin A and Hemoglobin S [typically in ~ 60:40 ratio]

Take home: When indicating the dominance/recessiveness of alleles, it is

important to also indicate the phenotype to which you are referring!!

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Variable penetrance and variable expressivity1

Variable expressivity of the Ayallele in mice:

Different alleles may show

variable penetrance and/orvariable expressivity. This

depends to a large degree on

the particular allele, but isalso influenced by non-allelic

modifying genes. Although

this can complicate genetic

analysis, it does open up newresearch to identify the

modifying genes and

determine how they influenceexpression of original gene.

Environment may also play a

role in some cases.

1Unfortunately, these terms are often used interchangeably among non-geneticists, though each term has a

specific meaning.

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