© 2006 Jones & Bartlett Publishers Chapter 2 classical Mendelian genetics

© 2006 Jones & Bartlett Publishers

Chapter 2

classical Mendelian genetics

http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm

Gregor Mendel

monkgardenercareful observerexperimenter

http://biology.clc.uc.edu/Fankhauser/Travel/Berlin/for_web/Mendel_in_Brno.html

Mendel

at that time (1850’s-1860’s) it was thought that the traits passed on by the parents were blended in the offspring

red + white = ?pink

purple + white = purple or white

Mendel

parents contributed to offspring factorsfactors remained unchanged

set out to trace their movements

looked at phenotypeappearancelooked at ratio’s

Mendel

used peas

access to different varietiesusually self-pollinatedcould manipulate pollinization

female

male

Mendel

started with true-breeding plants

round round round

wrinkled wrinkled wrinkled

parent 1 parent 1 offspring

x

x

Mendel

started with true-breeding plants

seed shape round vs. wrinkledseed color yellow vs. green

Mendel

seed shape round vs. wrinkled

round x wrinkled

wrinkled x round

round

round

Ppollen Pflower F1


Fig. 2.2. Reciprocal crosses of true-breeding pea plants

Mendel

seed shape round vs. wrinkled

round x wrinkled

wrinkled x round

round

round

Ppollen Pflower F1

hybridsdominant vs. recessive


Fig. 2.3. Traits studied in peas by Mendel

dom

inan

tdo

min

ant

rece

ssiv

ere

cess

ive

F1 cross

roundhybrid

roundhybrid

X

?


Fig. 2.7.

F1 cross

roundhybrid

roundhybrid

X

each hybrid has two parents

therefore

each hybrid has two “factors”

F1 cross

roundhybrid

roundhybrid

X

To solve:

define terms (use consistently)

determine parent genotypedetermine gamete genotypePunnett square

To solve:

R = round

? = wrinkled

capital letter=dominant

lower case letter-recessive

r

define terms

R and r are alleles:alternative forms of a gene

To solve:

original cross:

round vs. wrinkled(pure-breeding)

RR

determine parent genotypes

parents: rr

(homozygous)

To solve:

original cross:

round vs. wrinkled(pure-breeding)

RR rr

determine gamete genotypes

R or R r or r

parents:

gametes:

To solve: Punnett square

R R

Rr Rr

Rr Rr

r

r

all F1 would be Rr

(heterozygous)

F1 cross

roundhybrid

roundhybrid

X

RrRrparents:

gametes: R or r R or r

back to the

Punnett square

R r

RR Rr

Rr rr

R

r

What would the ratio of round to wrinkled be?

#1


Table 2.1. Results of Mendel’s monohybrid experiments

Mendel’s first law……The Law of Segregation

The two “factors” in the adult separate from each other during the production of the gametes.

homologous pairs of chromosomes separate during meiosis I.

#2

A problem:

Suppose you have a plant with purple flowers, but unknown ancestry. What is its’ genotype?

Is it homozygous dominantor

heterozygous?

PP

Pp

2 P’s or not…..

A solution:

cross it with homozygous recessive

P? x pp

do it!

A solution:

if PP x pp

if Pp x pp

all offspring would be Pp(heterozygous)(purple flowers)

some offspring would be Pp

some offspring would be pp(purple)

(white)

The solution:

Geneticists call this a “test cross”(pg. 44)

pause for ?

#3

Mendel also looked at two traits at once

round vs wrinkledyellow vs green

RR, rrYY, yy

pure-breeding round and yellow

pure-breeding green and wrinkled

pure-breeding round and yellow

pure-breeding green and wrinkled

RRYY rryyparentalgenotype

gametesgenotype

? ?RY ry

RY

ry

RrYy

all heterozygous; phenotype =?

all F1 were round, yellow hybrids RrYy(dihybrids)

RrYy RrYyparentalgenotype

gametesgenotype

? ?

Mendel then did a dihybrid cross (F1 cross)

metaphase I cell

R

yr

Y R

r

y

Y

splitclass

ryRY

ry

RY RRYY

RrYy

RrYy

rryy

rYRy

rY

Ry RRyy

RrYy

RrYy

rrYY

3 round, yellow1 wrinkled, green

1 round, green2 round, yellow1 wrinkled, yellow

round, yellowround, greenwrinkled, yellowwrinkled, green

315108101 32556

9331

Mendel’s results:

#4

(dihybrid cross)


Fig. 2.11. Independent segregation of the Ww and Gg allele pairs

Fig. 2.12. Dihybrid cross

Mendel’s second law……The Law of Independent Assortment (in modern language)

How a pair of homologous chromosomes align at Metaphase I is independent of how all the other pairs of chromosomes align at metaphase I

#5

metaphase I cell

yr

R Y R

r

y

Y

RY

ry

Ry

rY

2.4 probability

a fraction (ratio; like 1/4)

between 0

and 1

will never happen

will always happen

Y y

YY Yy

Yy yy

Y

y

Yy x Yyyellow yellow


2001 green6022 yellow

2001 (6022 + 2001)

= 1 4.01

Y y

YY Yy

Yy yy

Y

yWhat is the probability of gettingplants with green seed ?


What is the probability of getting plants with yellow peas?

YYYy

Y y

YY Yy

Yy yy

Y

y

1/42/4

prob [YY or Yy] = prob [YY] + prob [Yy]

= 1/4 + 2/4 = 3/4

prob [YY] + prob [Yy]

#6

RrYy x RrYy

What is the probability of getting plants with round, yellow peas?

Y y

YY Yy

Yy yy

Y

y

prob [round and yellow] = prob [round] x prob [yellow]

= 3/4 x 3/4 = 9/16

R r

RR Rr

Rr rr

R

r

9/16prob [round] x prob [yellow]

#7

YY or Yyand

RR or Rr

3/4

3/4

?

2.6 dominance ?

RR

Rr

R gene codes forstarch branching enzyme 1 (SBEI)

same phenotype ?


Fig. 2.20. Three attributes of phenotype affected by Mendel's alleles W and w, which determine round versus wrinkled seeds.

dominance is not necessarily all or none

RR and Rr same phenotype in peas, but really different at the molecular level

Are there some cases wherehomozygous dominant (RR)

looks different thanheterozygous (Rr) ?


Fig. 2.21. Incomplete dominance in snapdragons.

Incomplete dominance

When the heterozygote is intermediate between the homozygous phenotypes

Seen with traits that are quantitative(can be measured on a continuous scale)

as opposed to a discrete trait(appears to be all or none)

Blood typing (ABO)

Codominance

both traits are expressed

Human blood types:

AB

ABO

Blood typing (ABO)

IA

IB

i

three alleles (multiple alleles)

make “A” carbohydratemake “B” carbohydratemake neither carbohydrate

Fig. 2.22. The ABO antigens on the surface of human red blood cells are carbohydrates.

Blood typing (ABO)

Human blood types:

AB

ABO

phenotype: genotype:

IAIA, IAiIBIB, IBiIAIB

iicodominant

Blood typing (ABO)

Our immune system makes proteins

called antibodiesto attack foreign molecules

called antigens.

Blood typing (ABO)

For someone with type A blood (someone with the IA allele):

A carbohydrate “self”

B carbohydrate “non-self” or foreign antigen

Blood typing (ABO)

For someone with type B blood (someone with the IB allele):

A carbohydrate “non-self” or foreign antigen

B carbohydrate “self”

Blood typing (ABO)

For someone with type O blood (someone with ii alleles):

A carbohydrate “non-self” or foreign antigen

B carbohydrate “non-self” or foreign antigen

Blood typing (ABO)

For someone with type AB blood (someone with IA and IB alleles):

A carbohydrate “self”

B carbohydrate “self”

Blood typing (ABO)

What kind of antibodies would they make?

TypeABABO

- B antibodies- A antibodies- neither A nor B antibodies- A and B antibodies


Table 2.3. Genetic control of the ABO blood groups

AB universal recipientO universal donor


Fig. 2.23. Antibody against type-A antigen will agglutinate red blood cells carrying the type-A antigen.

Other reasons why Mendel rules aren’t always observed

incomplete dominancemultiple alleles

variable expressivitysame mutation-different results

penetrancecomplete 100%incomplete < 100%

polygenictraits ?

(phenotype = expected)

lung cancer

example

2.7 Epistasis

e.g.,When the expected 9:3:3:1 ratioof a dihybrid cross is altered

non-allelic genes interacting to affect the same trait

Merriam Webster: suppression of the effectof a gene by a nonallelic gene

2.7 Epistasis

C purple flowersc white flowers

plants with C- and P- genotypes havepurple flowers (wt)

plants with cc or pp have ? flowers

Given:

P purple flowersp white flowers

Two different genes C and P for flower color

white flowers

2.7 Epistasis aside:

How could that happen?(hypothetically)

A B C D purple pigment

mutation#1

mutation#3

E1 E2 E3 E4

2.7 Epistasis

CC pp x cc PP

phenotype

gametegenotype

F1 genotypeF1 phenotype

white flowers

Cp cP

CcPpall purple


Fig. 2.24. A cross showing epistasis in the determination of flower color in peas.

F1 crosswhat gametes ?

CP Cp cP cp

CP

Cp

cP

cp

9:7 ratiopurple to white

epistasis

When the expected 9:3:3:1 ratio of a dihybrid cross is altered

CCPP CCPp CcPP CcPp

CCPc CCpp CcPp Ccpp

CcPP CcPp ccPP ccPp

CcPp Ccpp ccPp ccpp


Fig. 2.25. Modified F2

dihybrid ratios.

2.8 Complementation

mutations c and p show complementation

CCpp x ccPPwhite white

CcPppurple


Fig. 2.26. Complementation reveals whether two recessive mutations are alleles of different genes.

2.8 Complementation

means that mutations affect different genes

means that mutations affect the same gene

Lack of complementation


Fig. 2.27. Results of complementation tests among six mutant strains of peas.

Fig. 2.28. A method for interpreting results of complementation tests.

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© 2006 Jones & Bartlett Publishers Chapter 2 classical Mendelian genetics