GENE INTERACTIONS ¾ A (AA, Aa) - PowerPoint Presentation

Slide1

GENE INTERACTIONS

Slide2

¾ A

(AA, Aa)

A B: ¾ x ¾ = 9/16

A b: ¾ x ¼ = 3/16

a B: ¼ x ¾ = 3/16

a b: ¼ x ¼ = 1/16

GENE INTERACTIONS

Consider two independent genes, A and B; two heterozygous individuals for both genes are crossed. Complete the scheme below:

AaBb X AaBb

Phenotypes and frequencies for gene APhenotypes and frequencies for gene BPhenotypes and frequencies for genes Aand B

¼ a

(aa)

¾ B

(BB, Bb)

¼ b

(bb)

¾ B

(BB, Bb)

¼ b

(bb)

Slide3

Because of the gene interaction phenomenon, some phenotypic classes are reduced in number, as two or more classes may display the same phenotype.

Which are the phenotypic classes of an F2 in which:

a) the dominant allele of a gene (A) masks the effect of the other gene (B)?

9/16 AB

3/16 Ab

3/16 aB

1/16 ab

9/16+3/16=

12/16

12:3:1b) the recessive homozygote (b) masks the effect of the other gene?

9/16 AB3/16 Ab

3/16 aB1/16 ab

3/16+1/16=

4/169:4:3

C)the phenotype due to the dominant allele of a gene is not observable from the phenotype due to the dominant allele of the other gene? Distinguish the two situations: when there is addictive effect and when there isn’t.

9/16 AB3/16 Ab

3/16 aB1/16 ab

3/16+3/16=

6/169:6:1

9/16 AB

3/16 Ab3/16 aB1/16 ab

9/16 + 3/16+3/16=

15/1615:1

Slide4

The

A gene encodes the enzyme α, the independent gene

B

encodes the enzyme β; these enzymes are both involved in the following metabolic pathway for cyanide production. The recessive alleles,

a and b, encodes for inactive enzymes.

a) Identify the genotypes of P and F1 individuals and determine which phenotypic classes are expected in F2 and with which frequencies.

a

b

S1 S2

cyanide

P

F1

Not producing plant X not producing plant

Producing plants

X

The parental plants don’t produce cyanide but ,after the cross, F1 plants produce cyanide. In parental plants one of two genes is homozygous recessive (inactive gene), the other gene is homozygous dominant (active enzyme)

Since the F1 plants are producing plants, the gene homozygous recessive are different between the parents.

(P1) AA bb X (P2) aa BB(F1) Aa Bb

Slide5

b)

b) what’s the name of genes that interact in the way showed here?

c)

which phenotypic classes are expected from the backcross of an F1 plant with the recessive homozygote?

Complementary Genes

, these genes act on the same metabolic chain.

(F1) Aa Bb x Aa Bb

(F2) 9:3:3:1

9/16 AB3/16 Ab

3/16 aB1/16 ab9/16 Producing plants

3/16 + 3/16 + 1/16= 7/16 Not producing plants Aa Bb x aa bb

1/4 AB1/4 Ab

1/4 aB1/4 ab

1/4 Producing Plants

¼ + ¼ + ¼ = ¾ Not Producing plants

Slide6

The two independent genes P and C control two different consecutive steps in the same metabolic pathway.

a) PPcc

P

C

S1 S2

Yellow fruit

The dominant alleles

P

and

C

encode for active enzymes, while the recessive alleles,

p

and

c

, encode for inactive enzymes. Which phenotype is expected for individuals with the following genotype (intermediate substrates are colourless):This plant produces the enzyme encoded by gene P, but not produces the other one (gene C). Metabolic chain is blocked at step S2

. Phenotype: white melonb) PpCcThe plant has both enzymes thus the metabolic chain works until the end.

Phenotype: yellow melon. c) ppCC

The plant hasn’t the enzyme encoded by gene P. It produces the enzyme encoded by the gene C but it doesn’t convert S1 in S2. Phenotype white melon.d) Ppcc

The plant produces the first enzyme but not the second one. Then the metabolic chain is blocked at S1 step. Phenotype white melon.

Slide7

The two independent genes, A e E, that originate through duplication, regulate the same step in the metabolic pathway that leads to the synthesis of a pigment.

The colourless phenotype can be seen only in the recessive homozygote state for both genes.

Aa Ee x Aa Ee

9/16 AE

3/16 Ae

3/16 aE

1/16 ae

1/16 white fur

9/16 + 3/16 + 3/16= 15/16 brown fur

Allele A produces enzyme

a

that converts S1 in pigment; allele a does not produce pigment.

Allele E produces enzyme e that converts S1 in pigment; allele e does not produce pigment.

Enzima a

S1 pigment Enzima e

Which phenotypic ratio is expected from the cross between two di-hybrids?

Slide8

Two pure lines of pepper are crossed: first line with Red fruits and the second one with orange fruits. The plants of F1 produce only red peppers, but crossing two individuals of F1 we obtain:

192 RED FRUITS, 47 ORANGE FRUITS, 14 WHITE FRUITS

Pure lines = they pass the same alleles to all

offsprings

and so they are homozygous

We know that the colour fruit is controlled by 2 genes (A and B), which kind of interaction is present? Determine the genotypes and hypothesize a possible action on metabolic chain.

(P1) RED FRUITS X (P2) ORANGE FRUITS

(F1) RED FRUITSF2:192 red fruits 47 orange fruits

14 white fruits When the plant has an allele of A the phenotype related to gene B is not shown. Phenotypes AB and Ab produce red fruitsWhen A is homozygous recessive, the plant show the B phenotype: if it has dominant allele B the fruits will be orange, if it is homozygous bb the fruits will be white.AA bbaa BB

Aa Bb9/16 AB + 3/16 Ab

3/16 aB1/16 ab192+47+14= 253 (:16= 15.8)

The dominant allele A is epistatic on gene B

Slide9

Phenotypes

Xo

H

Xa

(

Xo

–

Xa

)2: Xac2Red

19212/16189,755,06189,750,03Orange473/1647,430,1847,430,01

White141/1615,813,2715,81

0,21Total25316/16253

0,25We have to compare expected individuals (epistasis hypothesis) with observed individuals.

Degrees of freedom = 3-1 = 2c2 = 0,25P =80 - 90%Conclusion:

Hypothesis accepted

Slide10

Slide11

FILE 8:

POINT MUTATIONS

Slide12

GENETIC CODE

Slide13

1.

In a mRNA sequence (wt) there is a triplet UUU. After a mutation, the triplet changes in UUA. What kind of mutation happened and which effects have the mutation on the protein encoded by the gene?

To switch from UUU to UUA is necessary a mutation in the DNA

From

T T

T to T T A

AAA A A

TThis is a Transversion: it refers to the substitution of a purine for a

pyrimidine or vice versa, in deoxyribonucleic acid (DNA).

The aminoacid Phenylalanine encodes by UUU will be substitute with Leucine (UUA). This is a Missense mutation. The effect on the protein function is not predictable.

Slide14

2.

This is a sequence of wt mRNA

5’- AUG AGA CCC ACC….

What kind of effects has a mutation the substitute the fifth base from G to U?

5’– AUG AUA CCC ACC…

In the second triplet the mutation changes the codon and the aminoacid encoded:

Arg

(Encoded by

AGA) to Ile (encoded by AUA).What kind of effects has a deletion of the sixth base? 5’– AUG AUA CCC ACC…

Compare the two previous mutations:The deletion causes a frameshift of the codon code. All the aminoacids after will be different.

First case: Probably the protein will be active (Missense mutation)Second case: the protein is totally different from the original protein and it will probably not be active. (FrameShift Mutation).

Slide15

3.

A nucleotide sequence below

Knowing that the transcription of this sequence is from left to right

(5’->3’)

, write the resulting mRNA sequence:

b) Knowing that this mRNA sequence contains the translation start codon, identify the starting codon and indicate the amino acid sequence of the resulting peptide.

5 10 15 20 25 30 35

5'

ATTCGATGGG

A

TGGCAG

T

G

C

CAAAGTGGTGATGGC

3'

3'

TAAGCTACCC

T

ACCGTC

A

C

G

GTTTCACCACTAC

CG

5'

5’ AUUCGAUGGGAUGGCAGUGCCAAAGUGGUGAUGGC

Met- Gly- Trp-Gln-Cys- Gln- Ser-Gly- Asp- Gly

Slide16

5'

ATTCGATGGG

A

TGGCAG

T

G

C

CAAAGTGGTGATGGC

3'

3'

TAAGCTACCC

T

ACCGTC

A

C

G

GTTTCACCACTAC

CG

5'

C) Find which consequences will have on the amino acid sequence:

- a transition of the T base pair in position 18;

In the DNA sequence: switch from TA to CG;

In the mRNA sequence there is a switch between U and C thus between the triplet UGC (Cys) to

C

GC (Arg) that causes a MIS-SENSE mutation.

Slide17

5'

ATTCGATGGG

A

TGGCAG

T

G

C

CAAAGTGGTGATGGC

3'

3'

TAAGCTACCC

T

ACCGTC

A

C

G

GTTTCACCACTAC

CG

5'

- a transversion of the CG base pair in position 20;

Case 1: from CG to GC

Case 2: from CG to AT

mRNA: switch between C to G

From UGC (

Cys

) to UG

G

(

Trp

)

MISSENSE MUTATION: protein with one different

aminoacid

mRNA: switch between C to A

From UGC (Cys) to UG

A

(STOP)

NON SENSE MUTATION

thus a truncated protein

Slide18

5'

ATTCGATGGG

A

TGGCAG

T

G

C

CAAAGTGGTGATGGC

3'

3'

TAAGCTACCC

T

ACCGTC

A

C

G

GTTTCACCACTAC

CG

5'

- an insertion of a base pair after pair 11 (AT)

+1

AUG GGA

N

UG GCA GUG CCA AAG UGG UGA UGG C

Met-Gly-

aaX-Ala-Val-Pro-Lys-Trp-

Stop

After the insertion all the

aminoacids

will be different: frame-shift mutation.

The frame-shift creates a stop codon.

Slide19

+1

Met-Gly-

aaX-Glu

-Cys-Gln-Ser-Gly-Asp-Gly

d) How can we abolish the insertion mutation in position 11?

AUG-GGA-

N

UG-GAG-UGC-CAA-AGU-GGU-GAU-GGC

The second mutation restore the correct reading frame.

Only the aminoacids located between the two mutations will be different.

If in a position near the first insertion a deletion happens, we can restore the correct frame of aminoacids. For example in position 15 (intragenic suppression-suppressor mutation)

DNA with insertion and subsequent suppressor mutation:5’ ATTCGATGGGANTGGAGTGCCAAAGTGGTGATGGC 3’

3’ TAAGCTACCCTNACCTCACGGTTTCACCACTACCG 5’

Slide20

This is a

polypeptidic

sequence of a protein. Wild type and mutant sequences are compared. What type of mutations did happen?

WT :

Met-

Arg-Phe-Thr

…Mutant 1: Met-Ile-

Phe-

Thr…Mutant 2: Met-Ser-Ile-TyrWe compare mutant 1 with the wt: the second amino acid is different Arg could be encoded by CGU, CGC, CGA, CGG, AGA, AGG

Ile could be encoded by AUU, AUC, AUA Phe could be encoded by UUU/UUC and Thr by ACU/ACC/ACA/ACG, The possible DNA sequence will be:Sequence of wt: AUG – AGA – UUPy – ACN Py=C o TSequence of mutant 1: AUG – AUA – UUPy – ACN -

It is probable that the codon encoding Arg could be AGA, and a mutation occurred originating AUA

Slide21

WT :

Met-

Arg

-

Phe-Thr

…

Mutant 2: Met-Ser-Ile-Tyr

We compare mutant 2 with wt: all the amicoacids after the Methionine are different.

A frameshift mutation caused by an insertion. AUG – AGA – UUPy – ACN –Ser is encoded by UCN or AGU-AGC , We can hypotesize the sequence:AUG –AGN–AUU-PyAC-With N

=U/CThe third codon AUU = IleThe fourth codon will be UAC = Tyr

Slide22

5-

An Escherichia coli mutant auxotroph for tryptophan (Trp-) has an amino acid substitution in tryptophan-synthetase: Glycine at position 210 is replaced by an Arginine.

On the basis of the genetic code, find which kind of mutation (on the DNA) you think has caused the amino acid replacement

Gly Codons : GGU

GGC

GGA GGG

Arg Codons : CGU CGC CGA

CGG AGA AGG

We can hypothesize a single base substitutionFirst base G could switch to C or AGGU->CGU; GGC->CGC; GGA->CGA;

GGG->CGG;GGA->AGA; GGG->AGG

Slide23

In the Escherichia coli

metA

gene a base substitution occurred. Because of this mutation, in the mRNA a UAA codon is present inside the gene.

- Which consequence will this mutation have on protein synthesis?

The triplet UAA is a stop codon,

Thus the protein synthesis will be interrupted.

Slide24

Exon

INTRON

mRNA

The primary transcript of chicken ovalbumin RNA is composed by 7 introns (white) and 8 exons (black):

If the Ovoalbumin DNA is isolated, denaturated and hybridized to its cytoplasmic mRNA, which kind of structure do we expect?

Structure with loops corresponding to introns.

Slide25

if a deletion of a base pair occurs in the middle of the second intron, which will be the likely effects on the resulting polypeptide?

if a deletion of a base pair occurs in the middle of the first exon, which will be the likely effects on the resulting polypeptide?

We don’t have any effect if the mutation is not in the splicing site.

We have a frame-shift effect or a truncated protein.

Slide26

FILE 8

Slide27

A man has the chromosome 21 translocated on the 14. Draw the karyotype (only the chromosomes involved in the mutation).

What kind of gametes will be produced by this person?

Which will be the consequences on the progeny, if this man has a child together with a normal woman?

14

21

14-21

14

21

14-21

14

21

14-21

14

21

14-21

Pairing of homologous chromosomes during meiosis

Karyotype

GAMETES

Slide28

14

21

14-21

2

chromosomes

14, 1

chromosomes

21

14

21

14-21

3

chromosomes

14, 2

chromosomes

21

1

chromosome

14, 2

chromosomes

21

14

21

14-21

2 chromosomes

14, 2 chromosomes

21

2 chromosomes

14, 2 chromosomes

21

2 chromosomes 14, DOWN 3 chromosomes 21 SYNDROME

Parent with translocation

Normal Parent

Normal Gamete

Gametes (First parent)

21 Trisomy : Down’s Syndrome

21 Monosomy : not compatible with life

14 Trisomy : not compatible with life

Monosomy 14 n:ot compatible with life

Zigote healthy carrier: normal phenotype

Normal Zigote: normal phenotype

NOT VITAL

NOT VITAL

NOT VITAL

CARRIER

TRASLOCATION

NORMAL

Slide29

1/6+1/6+1/6: ZIGOTEs are not compatible with life

1/6 1/6 1/6

Which will be the consequences on the progeny, if this man has a child together with a normal woman?

Slide30

2-

In humans trisomy of chromosome 21 is responsible of the Down syndrome.

- which gametes originated an affected person?

draw a scheme of the meiotic stages that can give rise to the mutated gamete and indicate the name of this process.

Chromosomes are distributed to gametes incorrectly

The gametes either are missing or have an extra chromosome 21It is caused by the NONDISJUNCTION of chromosome 21 during meiosis (homologous chromosomes or sister chromatids).

Slide31

3.

A man carries a heterozygous paracentric inversion.

A B C D

E F G H

a b c d

g f e h

Draw a scheme of homologous chromosomal pairing during meiosis

Which gametes are produced (in particular which gametes are missing)? Explain why.

A B C D

a b c d

g

f

e

G

F

E

h

H

Draw only one cromatide for each chromosome.

A B C D

E F G H

a b c d

g f e h

Parental gametes

Slide32

Does the presence of the mutation change the fertility of this man?

Effect: a DICENTRIC CHROMOSOME (TWO CENTROMERES) and a ACENTRIC FRAGMENT CHROMOSOME (lost).

A B C D

a b c d

g

f

e

G

F

E

h

H

No, if inversion is not extended and is limited to a small part of chromosome.

RECOMBINATION

(CROSSING-OVER)

A B C D

E f g

h

e F G H

d c b a

RECOMBINANT GAMETES

We will have PARENTAL GAMETES but we don’t have vital recombinant gametes produced by a crossing over happens in the inverted region

Slide33

AA (1/2)

A (1/2)

BB (1/2)

B (1/2)

BB (1/2)

B (1/2)

CC (1/2)C (1/2)

CC (1/2)C (1/2)CC (1/2)C (1/2)CC(1/2)C (1/2)

AABBCC (1/8)AABBC (1/8)AABCC (1/8)AABC (1/8)ABBCC (1/8)ABBC (1/8)ABCC (1/8)ABC (1/8)

In a triploid cell which gametes are produced?

2/8 are gametes that could generate

an individual 6/8 are not compatible with life

6 - how can a triploid organism originate?

Draw a scheme of meiosis process in a triploid cell with n = 3.

From the cross between gamete n + gamete 2n.

Chromosomes A, B and C

gamete 2n= AABBCC

gamete n = ABCindividual 3n=AAABBBCCC

Slide34

MUTANTS in MICROORGANISMS

Slide35

The reproduction of the bacterium

Escherichia coli

is achieved by binary fission, after his genome replication. Complete the following scheme representing the chromosome in various stages.

Bacteria has one chromosome and don’t have nucleus.

Mitosis is performed in order to add cells to a population

Slide36

A prototroph bacterium can grow on a minimum media, composed by inorganic minerals and containing an

organic source of carbon

. Glucose is the most simple source of carbon.

Alternative carbon sources can be used by wild bacteria. Some mutants loose this ability

.The ability of bacteria to grow on different media is reported in the following table; identify their phenotypes.

STRAIN

Minimum media containing

PHENOTYPE

Glucose

Galactose

Lactose

Arabinosen. 1

++++

n. 2++

-+n.

3+--+

n. 4

+++-

wt

Lac-

Gal- Lac-Ara-

Slide37

A prototroph bacterium can grow on a minimum media, because it can synthesize what it needs. Mutants that aren’t able to grow on a minimum medium (auxotrophs) can be analysed on media enriched with different kinds of molecules, to determine their phenotypes.

The ability of bacteria to grow on different media is reported in the following table; identify their phenotype

STRAIN

Complete Media

Minimum media+ glucose and…

Phenotype

- -

arginine

pyrimidine

adenine

biotin

n. 1

+

-

+---

n. 2+-

-+--

n. 3+-

--+-

n. 4+

----+

Arg

-Pyr-

Ade-Bio-

Slide38

A mutant bacterium isn’t able to synthesize the prolin amino acid and to metabolize lactose. Indicate which molecules must be added to a minimum media to grow this mutant.

 

Minimum media + glucose and proline

30°C it grows

37°C it grows (optimal)

42°C it does not grow

An Escherichia coli mutant carries a temperature-sensitive mutation in the polymerase III gene that is necessary for DNA replication. At which temperature can this bacterium grow?

Slide39

Medium with streptomycin

Synthesize methionine

The wild-type

Escherichia coli

can’t grow on streptomycin-enriched media. If in a culture of 10 millions cells, there are about 100 streptomycin-resistant mutant cells, how can I select them (i.e. grow only the resistant cells)?

Two

Escherichia coli mutants display both the Met- phenotype, i.e. they aren’t able to....

The two mutations are located on different genes. How can you explain this situation?

S1 S2 methionine a b

Genes: met1 met2

Slide40

No because the pathways of galactose degradation is composed by different genes. If only one of the genes is inactive the galactose is not metabolized.

Two Escherichia coli mutants can’t grow on galactose medium if galactose is the unique source of Carbon.

Assign the phenotypes at the mutants

……………..

Are you sure that the mutations regard the same gene?

Gal

-

Slide41

FILE 10

Slide42

1.

In Escherichia coli, the lac (lactose) operon, is made of the following genes and sites. Specify what is the function of the ones indicated below:

- promoting site

- operator site

- repressor gene - structural genes.

PROMOTER

OPERATOR

STRUCTURAL GENESREPRESSOR

DNA region where RNA polymerase sits to start transcription. DNA locus where the repressor could bind to stop transcription.gene that encodes for a protein that negatively regulates transcription.

Genes that are usefull for a cellular function; for example metabolism of lactose.

Slide43

2.

What would be the result of a base substitution that inactivates the following genes:

Since the gene encodes for

b

-galattosidase enzyme, a mutation in this gene probably inactivates the function of the gene.

LacZ

-

LacI-

This gene encodes for the repressor of lactose operon. The mutation will have different effects depending on the protein domain where it occours:a) If the mutation inactivates the protein (frame-shift, stop codon, mis-sense), we have the absence of the repressor and constitutive transcription of the structural genes (recessive mutation LacI-)

b) If the mutation alters the allosteric domain of the protein where the inducer binds, we have constitutive repression of the structural genes because the repressor is bound to its site and is not influenced by the presence of lactose. (dominant mutation, LacIs)

Slide44

3.

What would happen if a base deletion occurs in the operator region?

OPERATOR is the DNA locus bound by the repressor to stop transcription.

After the mutation in the operator, the repressor could be unable to recognize the locus. In fact the repressor (protein) is able to recognize the Operator (sequence).

Thus, we have the constitutive expression of the genes. The mutant in operator constitutive lacO

c

(cis DOMINANT)

Cis indicates that the mutation acts on the nearest genes.

Slide45

4.

Which of the following genotypes will be able to produce β-galactosidase and/or permease

in the presence of lactose

?

b

gal perm.

…… …..

…… …..

…… ..….

…… ……

Lac I

+

Lac O

+

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

+

Lac O

+

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

+

Lac O

+

Lac Z

+

Lac Y

-

Lac A

+

Lac P

+

Lac I

+

Lac O

+

Lac Z

+

Lac Y

-

Lac A

+

Lac P

+

Lac I

-

Lac O

+

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

-

Lac O

+

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

+

Lac O

c

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

+

Lac O

c

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Genotype 1

Genotype 2

Genotype 3

Genotype 4

+ +

+ -

+ +

+ +

Genotypes 3 and 4 -> constitutive transcription of Lac operon (no repression)

Slide46

4.

If the

lactose is not present

, in which mutants the expression of genes change?

b

gal perm.

…… …..

…… …..

…… ..….

…… ……

Lac I

+

Lac O

+

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

+

Lac O

+

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

+

Lac O

+

Lac Z

+

Lac Y

-

Lac A

+

Lac P

+

Lac I

+

Lac O

+

Lac Z

+

Lac Y

-

Lac A

+

Lac P

+

Lac I

-

Lac O

+

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

-

Lac O

+

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

+

Lac O

c

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Lac I

+

Lac O

c

Lac Z

+

Lac Y

+

Lac A

+

Lac P

+

Genotype 1

Genotype 2

Genotype 3

Genotype 4

- -

- -

+ +

+ +

Genotypes 1 and 2 will not express the genes (REPRESSION)

Genotypes 3 and 4 will express constitutively the enzymes

Slide47

CIS dominant mutation: it expresses the dominant phenotype but it affects only the expression of genes on the same DNA molecule where the mutation occurs.

LacO

c

, affects only neighbouring genes (for example a plasmid)

To demonstrate it we construct an heterozygote (diploid) with:

The mutation lacZ- (

Bgal enzyme) located in cis to lacOc

no production of b-galattosidase

The gene lacZ+ in trans (on plasmid) with wt lacOPhenotype in absence of induction:mutation is cis-dominant -> no b-gal activitymutation is trans-dominant ->

b-gal activity5. What does it mean that the Oc mutation is dominant in cis? How can I demonstrate it? lacOc

lacZ-

lacO+lacZ+

lacIlacI

No lactose

Slide48

6.

What kind of phenotype will have a bacteria

LacI

+

O+ Z+ Y+

A+ carrying on a plasmid lacIs O

+ Z+ Y+ A

+ ? Which conditions can we use to analyse its phenotype?

The bacteria will be: LacI+ O+ Z+ Y+ A+ / LacIs O+ Z+ Y+

A+ heterozygote for gene LacI+ /LacISLacI+ encodes for the repressor that is able to bind lactose (induction and repression) and is able to recognise sequence of Operator (repression) LacIS encodes for a repressor with a mutation that unables the protein to interact with lactose (constitutive repression).LacIS

repressor is always bound to the Operator repressing the transcription of Operon Lac : the operon Lac genes are not transcribed and expressedIn order to analyse the phenotype I’ll grow the mutant in a condition where I can detect the expression of enzyme…with the presence of lactose (induction).We expect that the bacteria with the plasmid does not express the structural gene

 it does not grow if the unique source of carbon is Lactose. (trans dominant mutation)The wt bacteria (control) express the structural genes and grow on Lactose.

lacI

lacZ

lacI

SlacZ+

lacO

lacYlacA

lacO

lacYlacA

lactose

Slide49

I complement them: I produce bacteria partially diploid carrying both mutations: one on a plasmid with all the operon

Trp

, the other on the chromosome.

7.

Two bacteria have a

Trp- phenotype, then?

The Trp- bacteria are unable to synthesize Tryptophan. It needs Tryptophan to grow.I have to add Tryptophan in the medium.How can I verify whether the two mutations are in the same gene?

To analyse the phenotype I plate them on a medium

without TryptophanCASE 1: If bacteria grow, the two mutations complement each other, because they affect two different genes.CASE 2: If bacteria don’t grow, the two mutations do not complement each other, because they affect the same gene.

Slide50

POPULATION GENETICS

The study of the allele frequencies in a population

The Hardy Weinberg equilibrium

“Allele and genotype frequencies

in a population tend to remain constant in the absence of disturbing influences”

-non-random mating-mutations-selection-limited population size-random genetic driftConditions for HW equilibrium are never met in nature.

The equationsALLELES GENOTYPESp + q = 1 p

2 + 2pq + q2 =1

p=frequency of allele A p2=frequency of AA q=frequency of allele a q2=frequency of aa 2pq=frequency of Aa

Slide51

An Example

Assume a population in which 36% of the population are homozygous for a certain recessive allele a. Assume the population is at equilibrium.

What is the frequency of the recessive allele in this population?

36%

aaq

2=0.36q=f(a)= √ 0.36=0.6

What is the frequency of the dominant allele in this population?q =0.6p = f(A) =1-0.6=0.4What percentage of the population are homozygous for the dominant allele A?

p2=f(AA)=0.4 x 0.4=0.16 16%

What percentage of the population are heterozygous for the trait?2pq=f(Aa)=2 x 0.6 x 0.4 =0.48  48%Why do we have to start the problem with the percentage or numeber of the homozygous recessive in the population?Because only the recessive phenotype indicates only the homozygous recessive genotype…the dominant phenotype indicates heterozygous and homozygous

Slide52

0,1 0,4 0,5

In this case we know the number of heterozygous

We think about the number of

alleles

p =f (A) =

(2 x 100) + 400 = 0,3 2 x 1000q = f(a)=1 – 0,3 = 0,7

0,3 0,7

1 - Determine the gene and genotypic frequencies of the following populations:

Slide53

p =

(2 x 300) + 200

= 0,4

2 x 1000q = 1 – 0,4 = 0,6

0,3 0,2 0,5

0,4 0,6

1 - Determine the gene and genotypic frequencies of the following populations:

Slide54

p =

(2 x 700) + 200

= 0,8

2 x 1000q = 1 – 0,8 = 0,2

0,7 0,2 0,1

0,8 0,2

1 - Determine the gene and genotypic frequencies of the following populations:

(A(AA)+A(AA))+A(Aa)/(total alleles)

Slide55

2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.

Population 1

p = 0,3 q = 0,7

f(AA) = p

2 = 0,09

f(Aa) = 2pq = 0,42f(aa) = q2 = 0,49

Genotipi

Xo

Xa(Xo – Xa)2 : Xa

AA100901,11Aa4004200,95aa500490

0,20

c2 = 2,26

Degree of Freedom = 2(p,q)-1= 1P = 0,10-0,25Population is in equilibrium

Slide56

Population II

p = 0,4 q = 0,6

f(AA) = p

2 = 0,16

f(Aa) = 2pq = 0,48f(aa) = q2 = 0,36

Genotipi

Xo

Xa

(Xo – Xa)2 : XaAA300

160Aa200480aa500360

Expected and observed values are too different.

The population is not in equilibrium2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.

Slide57

Population

III

p = 0,8 q = 0,2

f(AA) = p2 = 0,64

f(Aa) = 2pq = 0,32f(aa) = q2 = 0,04

Genotipi

Xo

Xa

(Xo – Xa)2 : XaAA700640

5,6Aa20032045aa1004090

c2 =

140,6GL = 2– 1 = 1P <0,01The population is

not at equilibrium2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.

Slide58

3 -

The frequencies of

L

M and

LN alleles in a group of 200 black Americans were 0,8 e 0,2, respectively. Calculate the expected frequencies for individuals with M, N and MN blood group. 

If

the population is at

equilibrium:Group M = LM LM

= p2 = 0,8 x 0,8 = 0,64 x 200 = 128Group MN = LM LN = 2pq = 2 x 0,8 x 0,2 = 0,32 x 200 = 64Group N = LN LN = q2 = 0,2 x 0,2 = 0,04 x 200 = 8

Slide59

Population

I

f(aa) = 810/1000 = 0,81

Since

f(aa) = q2 f(a) = √0,81 = 0,9

f(A) 1 – 0,9 = 0,1

Population IIf(aa) = 360/1000 = 0,36f(a) = √0,36 = 0,6 f(A) 1 – 0,6 = 0,4

Population III

f(aa) = 490/1000 = 0,49f(a) = √0,49 = 0,7 f(A) 1 – 0,7 = 0,3 The individuals with phenotype A can be AA or

Aa.We are sure that individuals with phenotype a are: aaWe start from this information (aa) to figure out f(A)

Slide60

5 -

What is the expected frequency for dominant homozygous and heterozygous genotypes in a population at equilibrium in which the homozygous recessive genotype frequency is 0,09?

f(aa) = 0,09

f(a) = √0,09 = 0,3

f(A) = 1 – 0,3 = 0,7

f(AA) = 0,7 x 0,7 = 0,49f(Aa) = 2 x 0,7 x 0,3 = 0,42

Slide61

6 -

if in a population at equilibrium the frequency of Rh- phenotype is 0,0025, which is the expected frequency of heterozygous individuals?

f(Rh-) = 0,0025 = f(

dd

)

f(d) = √0,0025 = 0,05f(D) = 1 – 0,05 = 0,95f(Dd) = 2 x 0,95 x 0,05 = 0,095

Slide62

7

In Drosophila melanogaster the w recessive sex-linked allele, is responsible of the white colour of the eyes. In a population the frequency of this allele is 0,3. Which frequencies of male and female with white-eyes is expected if the population is at equilibrium?

The

frequency

of these two

alleles in the entire population are

𝑝 and 𝑞. Let's assume the locus is

present on the X chromosome. We

will assume that the allele frequency do not differ between males and females.

Male white eyes = w YFemale white eyes = w wFrequency:Males white eyes = f(w)=

q = 0,3Females white eyes = f(ww) = q2 = 0,3 x 0,3 = 0,09

If alleles are X-linked, females

may be homozygous or heterozygous, but males

carry only a single allele for each X-linked

locus. For X-linked alleles in females, the H-W

frequencies are the same as those for

autosomal loci. p2+2pq+q2=1In males,

however, the frequencies of the genotypes will be

p and q, the same as the

frequencies of the alleles in the population. p+q=1

Slide63

8 -

Daltonism is due to a sex-linked recessive allele. In a population the

daltonic

male frequency is 0,1; which is the one of

daltonic females?

Daltonic MALE = d Y (XY) hemizygote freq

dY = freq allele d

Daltonic FEMALE = d d (XX)

Frequency Daltonic Males = 0,1Frequency allele d in MALES = 0,1Daltonic FEMALES = f(d d) = 0,1 x 0,1 = 0,01Healthy carriers= 2*0.1*0.9=0.18

Slide64

relative Fitness (W) 2/2 = 1 2/2 = 1 1/2= 0,5

The selection is against the phenotype homozygous recessive

Fitness involves the ability of organisms to survive and reproduce in the environment in which they find themselves . The consequence of this survival and reproduction is that organisms contribute genes to the next generation.

Fitness= Its lifetime reproductive success

absolute Fitness 200/100 = 2 400/200 = 2 100/100 = 1

Slide65

10 -

A population is composed of: AA 250, Aa

500,

aa 250 individuals, when a selection factor acts against the

homozygote recessive and its fitness decreases to 0. The variation of genic frequency (selection effect) could be figured out taking into consideration the contribution of the gametes of different genotypes in the next generation.

Consider the previous population, try to figure out the variation of genetic frequency of a after a generation of selection.

1 

1 

0 

 250/1000

= 0,25

 

500

/1000 = 0,5

 

250

/1000 = 0,25

 1 x 0,25 = 0,25

1 x

0,5 =

0,5

 

 0,25 x 0 = 0

 [2(250

) + 500]/2(1000) = 0,5

[2(0

) + 0,5] /2(0,25 + 0,5 + 0) = 0,33

 q1 – q0 = 0,33 – 0,5 = -0,17

1000x2 total of alleles2x250 aa=250a+250a2(0.25+0.5) total of freq2x0 aa=0a+0a

As expected the frequency of a allele decreases

Slide66

11 – In a population there is a selection against the

homozygous recessive

with fitness = 0 but the genotypic frequencies are:

AA 0,64;

Aa

0,32;

aa 0,04,

Try to figure out the variation of genetic frequency of a after a generation of selection

1 1 0 0,64 0,32 0,04

0,64 x 1 = 0,64 0,32 x 1 = 0,32 0,04 x 0 = 0 [0,32 + 2(0,04)]/2(0,64 + 0,32 + 0,04) = 0,20 [0,32 + 2(0)]/2(0,64 + 0,32 + 0) = 0,16

q1 – q0 = 0,16 – 0,20 = -0,04Previous population Dq = -0,17 (> of q0 (0,5))

This population q0 = 0,2  With the same selection coefficient, the effect of selection against homozygous recessive change on varying of genic frequency

Slide67

12 – In a population the gene frequency of A is 0,3 and the frequency of a is 0,7, and there is a selection against the

heterozygous

(s =coefficient of selection=1-w= 0,4).

Try to figure out the variation of genetic frequency of a after a generation of selection

genotypes

AA

Aa

aa

fitness

 

 

 

Genotypic frequency

 

 

 

Gamete contribution

 

 

 

Frequency

of

q0

 

Frequency of

q1

 

D

q

 

1 1 – 0,4 = 0,6 1

p

2

= 0,3

2

= 0,09 2pq =

0,42

q

2

= 0,7

2

=

0,49

1 x 0,09 = 0,09 0,6 x 0,42 =

0,252

1 x 0,49 =

0,49

[

0,42

+ 2(

0.49

)]/2(0.09+0,42+0,49)=0,7

[

0,252

+ 2(

0,49

)]/2(0,09 + 0,252 + 0,49) = 0,74

q1 – q0 = 0,74 – 0,7 = 0,04

The recessive allele will be fixed and the dominant allele will be delete from the population

- What is the final result of this selection?

Slide68

genotypes

AA

Aa

aa

fitness

 

 

 

Genotypic frequency

 

 

 

Gametic

contribution

 

 

 

Frequency

of

q0

 

 

 

Frequency of q1

 

 

 

D

q

 

 

 

13 – The population is composed of: 640 AA, 320 Aa and 40 aa,

The fitness of homozygous dominant is 0,8 and the fitness of homozygous recessive is 0,1.

Try to figure out the variation of genetic frequency of after a generation of selection

0,8 1 0,1

640/1000=0,64 320/1000=0,32 40/1000=0,04

0,64 x 0,8=0,51 0,32 x 1 = 0,32 0,04 x 0,1=0,004

[320+2(40)]/2(1000) = 0,2

[0,32+2(0,004)]/2(0,51+0,32+0,004) = 0,19

0,19 – 0,20 = -0,01