Genetics Exam1 Pt2 Notes PDF

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Dr.Gilg's Exam 1 part 2 notes. Covers all lecture material including; How we can determine how many genes are affecting a trait looking at ratio and complementation test. Meiotic recombination and recombinant gametes. Independent assortment and crossing over. Calculating map unit distances for genes...

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How can we determine how many genes are affecting a trait?  Epistasismore than one gene is involved in determining phenotype 1) Look at Ratio 1 gene denominator is 4 2 genes denominator is 16 3 genes denominator is 64 -

This is difficult because a 9:7 ratio could possibly be a 1:1 *shrug*

2) Complementation test o Complementation – when a cross between 2 individuals that both show a recessive phenotype results in the production of offspring with the dominant phenotype

A = Red

a = white

P

AA x aa

aa x aa

F1

Aa (Red)

aa

F2

¾ A-

No complementation

¼ aa

Colorless 1 ----[A] --- > colorless 2 --- [B] --- > Red P

AABB (Red) x aabb (White)

F1

AaBb (Red)

AAbb (white) x aaBB (white) AaBb (Red)

Complementation

Can only get complementation if you are looking for a trait that is influenced by more than 1 gene Cannot get complementation by crossing two individuals’ that are homozygous recessive at the same gene

2cont.) Complementation Test – o

When all pairwise crosses are made among individuals that have the recessive phenotype

Ex. Harebell Flowers -

Most plants have Blue flowers but find 3 plants with white flowers o Plant 1 = X o Plant 2 = Y o Plant 3 = Z

P

Blue x X

Blue x Y

Blue x Z

F1

all blue

all blue

all blue  tells us that Blue is completely dominant to White

F2 ¾ Blue ¼ White ¾ Blue ¼ White

¾ Blue ¼ White 

Tells us X, Y, Z are all homozygous recessive for a single gene that influences flower color

Could they be recessive at different genes?? Are X,Y, and Z all homozygous recessive for the same gene? Or could there be more than one gene involved in flower color To solve this, we must do a complementation test -

We will cross X with Y, X with Z, and Y with Z to see if any of the crosses result in complementation

Complementation Test – Make all pair-wise crosses between individuals with recessive phenotype to determine if any of the crosses result in complementation.

X x Y = white Complementation. The only way that this can happen is X x Z = Blue  if X and Z are homozygous recessive for different genes Y x Z = Blue  Complementation.

-

X Y Z

Usually results will not look like this. Usually they will be put into a matrix X

Y

Z

W

W

B

W

B W

Bateson + Punnett -

-

Garden Peas 2 traits: o Flower Color (Purple or Red, Purple completely dominant) o Shape of Pollen Grain (Long or Round, Long completely dominant) Took True Breeding Purple, Long and crossed with True Breeding Red, Round

PPLL x ppll F1

PpLl  All Purple Long

F2

Expected: 9:3:3:1 9/16 Purple, Long 3/16 Purple, Round 3/16 Red, Long 1/16 Red, Round

F2

Observed: P-L-

4831

P-ll

390

ppL-

393

ppll

1338

Segregation is seen. But Independent Assortment is not seen.

They hypothesized that some linkage caused parental combinations to be seen more often Morgan – Drosophila melanogaster -

-

Found similar patterns to Bateson and Punnet in some traits in fruit flies Eye Color o Pr+: Purple o Pr: Red Wing Length Completely Dominant o Vg+: Long wings o Vg: vestigial wings

Are there times when genes are NOT independently assorted? Are some combinations of alleles found in the gametes more often than others? How can we determine what combinations of alleles are being carried by a gamete?? -

Needed to do a Test Cross

Test Cross: -

Mate an individual with a dominant phenotype with an individual that is homozygous recessive.

Aa:Bb x Aa:Bb  Could we determine from the offspring produces, which gametes produced them? (Let’s focus on one of the genotypes) A-BTest Cross:

ABxab

 Many possible gametes could make a A-B- offspring

ABxAb

Aa:Bb x aa:bb

ABxAb etc….

Gametes A:B

ab

 Aa:Bb

A:b

 Aa:bb

a:B

 aa:Bb

a:b

 aa:bb

Test cross allows us to determine the proportions of each gamete. Phenotypic ratio from a test cross will be the same as the ratio of the gametes being produced. So P

pr+pr+: vg+vg+ x

prpr: vgvg

F1

pr+pr: vg+vg x prprvgvg  Test Cross to determine the ratio of the gametes made by F1

Gametes

pr+vg+

 pr+: vg+

1339

pr+:vg

 pr+: vg

151

pr:vg+

 pr: vg+

154

pr:vg

 pr:vg

prvg

1195

2 potential explanations for lack of 1:1:1:1 ratio in test cross -

1. Parental combinations of alleles are found together in gametes of F1 more often than expected or 2. Dominant alleles are found with other dominant alleles (and recessive with recessive) in gametes more often than expected. Some strange force of attraction between them.

P

pr+pr+: vgvg

x

prpr:vg+vg+

F1

pr+pr : vg+vg

Gametes

pr+ : vg+

Hypothesis 1 supports the middle two gametes being found more often. Hypothesis 2 supports the top and bottom gametes being found more often in progeny

x

prpr:vgvg

(test cross)

pr : vg  157

pr+ : vg

 965

pr : vg+

 1067

pr : vg

 146

Hypothesis 1 is supported by the results of this test cross. Parental combinations of alleles are found together in gametes of F1 more often than expected.

How are parental combinations passed along together in most cases? -

Physical linkage of genes

**Each chromosome consists of a single molecule of DNA. (Draws parent from cross above making gametes  F1  gametes of F1 w/ NO crossing over) How can gametes with different combinations of alleles than the parental generation be formed by F1? Meiotic Recombination: -

Any process that results in the production of recombinant gametes

Recombinant gametes: -

Any gamete with a combination of alleles that differs from what the parental generation could produce.

-

P

AABB x aabb

Gametes

AB

ab

F1

Aa:Bb

Gametes

AB

 Parental

Ab

 Recombinant

aB

 Recombinant

ab

 Parental

Gametes can be defined as parental or recombinant.

Meiotic Recombination – 2 mechanisms 1) Independent Assortment – Will produce recombinant gametes at a frequency of 50% a. RF (recombinant frequency) = 0.5 [highest proportion of recombinance possible] i. (typically happens when 2 genes are on 2 different chromosomes. And predicts that they assort at equal frequencies) 2) Crossing Over – Homologous chromosomes can sometimes exchange parts. a. If the crossover event occurs between the genes of interest, then it will produce recombinant gametes for those genes i. (happens during prophase 1 of meiosis. Happens when 2 genes are on the SAME chromosome) b. Recombination event will ONLY occur if the chromosomes cross over in BETWEEN the two genes of interest. c. RF (recombinant frequency) = < 0.5 when genes are linked. Could only possibly be 0.5 if crossing over between genes of interest happened EVERY TIME. Which doesn’t happen. o

The distance of the genes on the chromosomes will dictate how often crossing over occurs  The further apart the two genes are, the more often recombination will occur.  This will allow Linkage Mapping.

o

Since recombinant frequency will increase with the distance between two genes, we can use this information to map the relative locations of genes on a chromosome

Locus (singular) – a physical location on a chromosome (can use locus and gene interchangeably) Loci (plural) Map Units (m.u.) or CentiMorgans (cM) -

Fun Fact: Only about 5% of DNA encodes for proteins (genes). The other 95% aren’t really “genes”

Unit of distance between two loci o only tells you relative distance, does not convert to any other unit of measurement m.u. = RF*100

Linkage Mapping – -

Must have RF data for all pairwise combinations of the loci you are investigating. Ex. o RF of A to B = 0.05 = 5 mu o RF of A to C = 0.01 = 10 mu

How far apart are B and C? A--5--B--5--C

or

B--5--A--10--C this is why we need ALL pairwise combinations.

Calculating RF for pairs of genes Address 2 questions: 1) Are the two genes linked? No, if RF = 0.5 Yes, if RF < 0.5 2) If Yes, how far apart are the loci? Calculate RF*100 = m.u. Must conduct test crosses to determine the RF between loci Example 3 Point Test Cross -

A test cross involving 3 loci

sc – thoracic bristles sc+ or sc ec – eye surface texture ec+ or ec vg – wing length vg+ or vg P – sc+sc+:ec+ec+:vg+vg+ F1

x

scsc:ecec:vgvg

sc+sc:ec+ec:vg+vg

x

scsc:ecec:vgvg (test cross)

RF =

¿ of recombinant gametes Data: for test cross

sc vs ec

sc ec vg

235

Parental

recombinants

sc+ ec+ vg+

241

sc+ec+

scec+

sc ec vg+

243

scec

sc+ec

sc+ ec+ vg

233

RF = (12+14+14+16)/ 1008 = 0.056

sc ec+ vg

12

sc and ec are linked and are 5.6 m.u. apart

sc+ ec vg+

14

sc ec+ vg+

14

ec vs vg

sc+ ec vg

16

Parental

recombinants

total

1008

ec+vg+

ecvg+

ecvg

ec+vg

-

sc vs vg Parental

recombinants

sc+vg+

scvg+

scvg

sc+vg

RF = (243+233+14+16)/ 1008 = 0.502 sc and ec are not linked

RF = (243+233+12+14)/ 1008 = 0.498 sc and ec are on the same chromosome. Linked. Is vg on the same chromosome? sc and ec are not linked o Can find out if we find linkage with ec  Not linked, therefore vg is on separate chromosome

Should be able to work it in reverse. Given RF make data for test cross

Loci: A and B are 20 m.u. apart P

AAbb

x

aaBB

F1

AaBb

Gametes

AB

= 0.1

Ab

= 0.4

aB

= 0.4

ab

= 0.1

RF = 0.2 = 20% recombination frequency PF = 0.8 = 80% recombination frequency What is the frequency of each gamete produced by the F1?

We could take the F1 and mate it with another F1 and use the proportions in order to determine phenotypic and genotypic ratios

RF = probability of a crossover F1

-

A------b

prob. of crossover between A+B = 0.2

a------B

prob. of no crossover is = 0.8

No crossover Ab aB = 0.8 (0.4 freq of each) Crossover AB ab = 0.2 (0.1 freq of each)

Can do with multiple genes A—0.1—B—0.2—C -- 0.15—D a-----------b----------c-------------d

RF=X A

Y B

C

RFA to C < X + Y -

RF between genes that are far apart will be an underestimate of the actual distance between them. o WHY? Long map distances are often underestimated due to the occurrence of double crossovers (DCO’s) DCO’s produce extra parental gametes, so RF will be an underestimate

Interference -

The degree to which one crossover inhibits the formation of a second crossover.

I = 1 – C.o.C. (Coefficient of Coincidence) C.o.C. =

Observed DCO ' s Expected DCO ' s

Expected freq. of DCO = RF1 x RF2 = (X)(Y) Observed freq. of DCO = Observed DCO/ Total

Mendelian Inheritance – -

Each parent contributes equally to the genotype Phenotype of individual is determined by its genotype  The combination of alleles inherited from its parents

3 types of Non-Mendelian Inheritance 1) Inheritance of Extranuclear genes o Genes located on chromosomes of mitochondria or chloroplasts 2) Maternal Effects o When the phenotype of an individual is not determined by its genotype, but by non-genetic influences of the mother 3) Parental Imprinting (Epigenetics) o When the phenotype is determined by the allele inherited from only one parent Extranuclear Genes -

-

-

mtDNA (mitochondria) and cpDNA (chloroplast) Mitochondrial + Chloroplast chromosomes – o Circular o No Histone proteins o Haploid o Bacterial ribosomes o Other similar processes that occur in bacteria – reproduce by binary fission, etc. Mitochondrial Genes o Contains most of the genes necessary to encode proteins involved in cellular respiration. o Encodes its own tRNA + rRNA molecules  So can translate its own proteins o (Mitochondrial ribosomes tend to be bigger than cp’s and has some more genes encoded) Chloroplast Genes o Encodes most of proteins involved in photosynthesis

-

o Encodes tRNA + rRNA to translate proteins. Inheritance of mtDNA and cpDNA o Vast majority of eukaryotes show Maternal Inheritance of mtDNA and cpDNA.

Why? -

In most organisms the mitochondria/chloroplast of sperm don’t enter cytoplasm of egg. What is this going to look like on a pedigree?? o Traits due to mtDNA will show mother with trait passing to all offspring, but only daughters will pass it to their offspring (all)

Exception: A few plants species show paternal inheritance of extranuclear genes -

Doubly uniparental inheritance (observed in some species of marine mussels) o Females – inherit mitochondria from mother  Pass to both daughters + sons o Males – inherit mtDNA from mother + father  Only pass to sons  But only pass the mtDNA they inherited from their fathers How?  Somatic Tissues in males are heteroplasmic (contain mitochondrial DNA from both parents in approx. equal proportions  Germ Tissues in males are homoplasmic – only contain the mtDNA from father

Maternal Effects -

-

When the phenotype of the individual is not determined by its genotype but due to non-genetic influences of mother. Sometimes environmental effects o Ex. Crack babies, fetal alcohol syndrome, other chemicals during development o Nutrition/Energy budget of the mother o Location of oviposition (in insects) or in other animals, where mother lays eggs Can be a genetic basis o In many cases the mother transcribes + translates genes of her somatic cells and the mRNA of proteins are transported into the egg where they guide early development.

Ex. : Snail Linnaea perega -

P F1

Direction of shell coiling is due to gene D (a single gene). D dominant – shells coil to the Right dd recessive – shells coil to the Left o The Genotype of the mother determines this! Not the genotype of the individual. o Mother produces proteins of D in her somatic cells o These are transported into egg + will guide shell development after fertilization DD (male) x dd (female) Dd  shell still coils to the left because the mother only produces recessive proteins (dd)

DD, Dd, dd  All coil to the Right because the mother is only producing the dom. proteins (Dd)

F2

Randomly mate F2 to produce F3 F3

¾ will coil Right

¼ will coil Left

*Be able to identify that something is off mendelian inheritance when in F2 you did not get a 3:1 ratio** *Know all of the other options*

Parental Imprinting -

When the phenotype of offspring is determined by the allele inherited from only one parent Maternal Imprinting – Allele from mother is inactive so only allele from father is expressed. Paternal Imprinting – Fathers allele is made inactive, only allele from mother is expressed. o Imprinting – to make inactive/shut off Fairly rare, very few genes do this. But those that do, do not switch back and forth between maternal and paternal, they strictly do one or the other.

Ex. In rats, Inheritance of igf2 – insulin-like growth factor 2. -

Shows maternal imprinting o Allele inherited from father determines the size.

Igf2+  normal growth Igf2-  dwarfism P

Igf2+igf2+ (male) x igf2-igf2- (female) igf2+igf2-  Normal growth

F1

Reciprocal cross P

Igf2+igf2+ (Female) x igf2-igf2- (male) igf2+igf2-  Dwarfism

F1

Regardless of what the fathers phenotype is, he can pass down either version of allele If heterozygous. How will you determine sex-linkage from paternal or maternal imprinting?? -

Sex linked will show different ratios in males and females. Imprinting will not show different ratios in males and females.

**Things to master for exam: Laws of probability, product rule + sum rule Law of Segregation – Why does it happen  Meiosis Law of Independent Assortment – Why does it happen  Meiosis The number of phenotypes in Data and the ratio can tell you a lot about the inheritance Be able to recognize patterns of all different types of inheritance....