-the segregation of chromosomes results in the segregation of alleles

-to assess the segregation of alleles, you must determine the genotype of the meiotic products

-if an allele confers a particular phenotype, the genotype can be inferred from the phenotype

*in a haploid life cycle, the haploid products of meiosis (non-transient) can be scored phenotypically

-they are simple... no hetero/homo, no dom/rec

*in a diploid lifecycle, the haploid products of meiosis are transient and cannot be scored phenotypically

-the diploid organism that results from their fusion can be scored phenotypically

-based on the phenotype of the diploid, one can infer the genotype of meiotic products

next level of complication

-in both diploid and haploid lifecycles, the products of multiple meioses are usually pooled

-therefore, the segregation of alleles within a single meiosis can only b inferred from the ratio of alleles within the pool

-some fungi produce tetrads, in which products of a single meiosis are not pooled, but rather are maintained in an ascus

-in other fungi, the tetrads are linear, meaning that they reflect the segregation of chromatids during meiosis (get octad due to post-meiotic mitosis) 

first division segregation = MI segregation

-a 4:4 ratio in spores, with differing alleles being separated by the first meiotic division

-indicates that before meiosis I, different alleles were on different homologues and therefore that no recombination occurred

second division segregation = MII segregation

-a diploid heterozygous parent undergoes meiosis

-results in a 2:2:2:2 ratio in spores, with differing alleles being separated by the second meiotic division

conclusion

-meiosis I does NOT separate different alleles, therefore different alleles must have been in the same homologue (chromatids joined by a centromere)

-suggests that crossover occurred b/w gene and centromere, so that the centromere links 2 non-identical chromatids 

-the frequency of crossing over b/w any two points on a chromosome is directly proportional to the distance b/w the two points

-expressed as:

distance b/w 2 points = % meiotic products showing recombination b/w two points

-in linear octads, the frequency of MII octads is a phenotypic indication that a recombination event has occurred b/w the gene and the centromere

-therefore, can be used to calculate distance b/w gene and centromere  (2 points)

-within each MII octad, spores = products of meiosis

-within these products of meiosies, only 1/2 show recombination b/w gene and centromere

-therefore, the distance from gene to centromere can be determined from…

[(1/2 MII octads)/(total octads scored)] * 100 

% = map units

*depending on how the spindle attaches to the centromere, 4 different but equally frequent MII patterns arise 

Mendel (experiments)

used peas:

1)cheap, small, easy

2)quick life span, many progeny

3)different characteristics available

4)cross or self pollination

5)diploid

Hypothesis 1

Crossing 2 parents /w distinct traits will result in offspring with intermediate traits

-generated true breeding lines (homozygous), purple and white flower

-expected blended phenotype

Conclusion: Purple Phenotype is dominant

2

Is the white characteristic still present (but masked) in purple F1?

-allowed F1 to self fertilize...

-created second filial (F2) generation

-got ratio about 3:1 phenotype (corresponds to 1:2:1 genotype)

Conclusion : white reappears in F2, therefore must be present but masked in F1 .... recessive trait

Next...

What accounts for the 3:1 ratio?

-generates F3 progeny from each phenotypic class in the F2

Hyp 3:

One copy of purple factor and one copy of white factor are in the F1 (p/w) and each parent passes only one (p or w) to progeny

 -he breeds F2 with itself

-pp=all purple

-pw = 3:1

-ww=all white

**difference b/w p and w traits is that they carry a discrete and different hereditary determinant (factors) ... alleles of a gene 

1. Difference in phenotype is due to discrete and different hereditary determinants or factors (factors = genes, different factors = alleles)

2. each plant has 2 copies of the factor (2 alleles) for each character

3.for each factor, one copy is dominant to the other (not entirely true)

4.Parents pass one copy of each factor to offspring

-one member of each gene segregates randomly into the gametes, so that each gamete has one allele

 (meiosis)

5.Union if gametes from each parent is independent of allele the gamete carries (fusion is random)

Mendel's first law:

law of equal segregation

-the two members of a gene pair (alleles) segregate equally into the gametes, so that 1/2 the gametes carry one allele, and 1/2 the gametes carry the other allele

-=segregation of homologous chromosomes 

Hypothesis:

Purple F1 is heterozygous, 2/3 of purple plants in the F2 are heterozygous

TEST CROSS

-an individual of an unkown genotype is crossed to a homozygous recessive individuals

-in a test cross, the tester (pp) contributes only recessive gametes therefore the progeny are a direct indications of the gametes produced by the tested individual

-homozygous recessive (tester) contributes nothing to decision

-if phenotypic ratio is 1:1, genotypic is also 1:1

-phenotype ratio reflects genotype of unknown .. can either be 2:0, 1:1, or 0:2

MENDEL's SECOND LAW (not related to this)

Law of Independent Assortment

-gametes assort independently

-random event = we can determine probabilities

F2 genotype ratio: 1PP:2Pp:1pp

-F2 phenotypic ratio: 3:1

-in many animals and some dioecious plants, sex is determined by the presence of particular chromosomes, the sex chromosome

-remaining, non-sex chromosomes are called autosomes

-the two sexes carry a different pair of sex chromosomes

-in most species, males carry an XY pair, while females carry an XX pair

-X and Y chromosomes can pair during meiosis because of the presence of small areas of homology, the pairing region

-remaining regions of the X and Y chromosomes carry different genes, with the Y chromosome carrying only a small number of genes

homogametic sex (females) produce gametes containing only X

heterogametic sex (males) produce gametes that contain both sexes (X and Y)

sex linkage: location of a gene on a sex chromosome

-gene is on one of the sex chromosomes, usually the X

-because the gene is on a sex chromosome, the phenotype associated with that gene will be associatd with one sex more frequently than the other

-

-there are way more genes on the X chromosome than the Y

-whatever guy has is expressed, while females have a backup

-prevalence on colorblindness exists on X-chromosome

**for any gene on the X-chromosome, the male is said to be hemizygous, since he only has one copy, it IS expressed phenotypically

-first recognized sex linkage is Drosophila in 1910

-used eye color .... w+ is red... wild type

-W is mutant... white eyed

Began with Mendelian experiments....

...red eyed female with white eyed male (both homo) yielded all red in F1 and 3:1 red:white in F2..

normal right?

BUT..

-when he did white female crossed with red male...

-yielded 1:1 ratio in F1

conclusion:

-reciprocal crosses give different results

***in reciprocal cross, white female is homo recessive!!!! (white... obviously)

hypothesis:

ene W must be on the X chromosome

KEY POINTS

- all females get Xw+ from father = all are red

-males receive Y from father that does not carry the w gene and therefore does not contribute to eye colour phenotype

-phenotype depends on the mother's contribution

-males get either Xw+ or Xw from mother (half red half white)

-male is called hemizygous (ie. XwY)

-has only 1 allele of the w gene, so cannot be homozygous or heterozygous

-whatever allel is present on the male' X chromosome is expressed phenotypically

-applies the principles of Mendelian genetics to humans, where controlled crosses are not possible, and existing family trees must be used

-random crosses

-small numbers, may not fit expected ratios (small df)

RULES
Autosomal recessive trait (ie.albinoism)

1.Phenotype appears with equal frequency in each sex

2.Two unaffected individuals produce both affected and unaffected children

3.Two affected individuals produce only affected children

assumptions.... if an allele is rare and recessive, an unaffected individual marrying into a family will most likely be homozygous for the dominant allele (we assume this)

***rare does not = recessive/dominant (can be rare and dominant, or rare and recessive)

Autosomal dominant trait

1. Phenotype appears with equal frequency in both sex

2.Affected individuals produce both kinds of children

3.Affected individuals in  every generation

4.unaffected individuals produce unaffected only

X-linked recessive trait

1.More males than females with the trait

2.No children of an affected male show the trait, but 0.5 of grandsons will

3.skipping generations = recessive

X-linked dominant trait

1.Affected males pass the trait on to their daughters but not their sons

2.Females married to unaffected males pass the trait on to 0.5 their sons and daughters

3.males receive from females

-both mitochondria and chloroplasts contain DNA
-during cell fusion in sexual reproduction, male and female cells contribute nuclear DNA equally, but the female contributes most of the cytoplasm

-cytoplasmic DNA comes only from the mother

-maternal inheritance

-sperm has virtually no cytoplasm

-cytoplasm of zygote comes from egg

MATERNAL INHERITANCE
-reciprocal crosses give different results when gene is in a organelle genome

-poky- mitochondrial gene, mutants are slow-growing (shown as brown cytoplasm)

Ad=nuclear gene

-gene is located in mitochondria

-/w poky male, all progeny are 'normal'

-look at figure 5-22

-in Aa, presence of A (functional enzyme) will override the presence of a non-functional enzyme, therefore A will be dominant to a

Aa = A phenotype

EXAMPLE

Phenotype depends on a threshold amount of protein

eyeless+(ey+)-required to form eyes in Drosphilia

-each ey+ allele makes 50 units protein/cell

-a wild type makes 100

CASE 1

-a cell required 80 units of functional Ey protein to form an eye

-mutant allele ey-1 makes 0 units (null)

-a homo mutant has no eye (0 units)

 -ey+/ey-1 will have no eyes, as the requird protein level is not met

CASE 2

-a cell requires 80 units of functional EY proein to form an eye

-mutant allele ey-2 makes 35 (leaky)

-a homo mutant has no eye (70 units)

-hetero however has eyes, as the required protein level is met

These cases have heterozygotes that cannot be distinguished from the homozygote dominant (either mutant or wild type)

-all or nothing

EXAMPLE
-phenotype varies with the concentration of protein

-True breeding red-flowered snap dragons are crossed with tru breeding white-flowered snapdragons

-Mendel's work predicts all red or all white offspring

-however, all offspring are pink

-when F1 is crossed, get 1/2 pink, 1/4 white(rr) and 1/4 red(RR)

Conclusion

-if the heterozygote has a phenotype distinct from either homozygote, the phenotypic ratio will be the same as the genotypic ratio

-two allele 'doses' produce the largest amount of transcript

-a single 'dose' results in less transcript

The heterozygote is intermediate b/w two homozygous individuals = incomplete dominance

-Incomplete dominance occurs if the heterozygote has a phenotype distinct from either homozygote, the phenotypic ratio will be the same as the genetic ratio

What if alleles produce proteins having different and independent functions?

Blood Groups

-blood is categorized into different types based on how blood reacts to a set of antibodies

-each antibody interacts with a particular antigen (protein) present on the cell surface

-the presence of such an interaction is recognized by blood clumping

-one antigen (protein) is produced by gene I

-3 alleles (I^A,I^B,I) which produce different antigens recognized by different antibodies

I^A reacts with A antibody

I^B reacts with B antibody

I reacts with no antibody (null allele)

I^AI^A or I^Ai = Antigen A = blood type A

I^BI^B or I^Bi = Anti B = b.t. B

I^AI^B = Anti A + B = b.t. AB

ii = no antigen = b.t. O

codominant = heterozygote has phenotypic characteristics of both homozygotes

therefore,

I^A and I^B are codominant to one another

-both are dominant to i

-sickle cell anemia, a tropical disease cause by a mutation in the gene coding for hemoglobin

-Hb^A/Hb^A = wild type, NO anemia

-Hb^S/Hb^S = ANEMIC

-Hb^A/Hb^S = blood cells normal except under low O2

DISEASE = complete dominance... only one genotype exhibits anemia

CELL SHAPE = incomplete dominance.... heterozygote is slightly sickle shaped

GEL MIGRATION = codominance... heterozygote exhibits characteristics of both

***If we consider different phenotypic characteristics, the dominance/recessive descriptions vary

Manx cat

-manx are tailless due to abnormal spinal chord development

-they breed to produce 2 manx cats and 1 tailed cat (2:1)

-not true breeding... is 1:2:1 with one group missing

-this is a Mendelian ratio, but one class is combined with another or missing (hypothesis)

-m is a lethal allele, and recessive homo dies

-to test, must cross 2 manx cats and look for lethality during embryogenesis

conclusion: M+ is necessary for spinal cord development

M+M have sufficient wild type protein to make most of the spinal cord, but the tail is absent

Question: If we consider 2 genes controlling two different characteristics, is their segregation independent or dependent of one another?

Prediction: consider the behaviour of chromosomes:

-if genes are on different chromosomes, we expect alleles of the genes to segregate independently of one another

-if genes are on the same chromosome, we expect tha the alleles that came from one parent will remain together unless a crossover occurs between them

Mechanisms that result in a recombinant

1)independent assortment of genes on different chromosomes (Mendel's 2nd law)

2)crossing over b/w genes on the same chromosome

A recombinant gamete is any gamete that has a genotype that differs from the genotypes of teh two haploid parents that contributed to the diploid meiocyte

A parental gamete is any gamete that has the same genotype as one of the two haploid parents that contributed to the diploid meiocyte.

How does the position of the genes within the genome afect the inheritane pattern?

Part 1: Inheritance of genes that are on different chromosomes (independent assortment)

Question: What proportion of each meiotic product will be generated, if two genes are on different chromosomes?

Mendel chose two characteristics, seed colour (yellow=Y or green=y) and seed shape (round=R and wrinkled=r), and analyzed their segregation in:

1)individuals differing in one of the characteristics 

2)individuals differing in both characteristics

-first determined that monohybrid cross yielded 'normal' segregation (ie. 3:1 in F2)

-next, he crossed plants that were different at two genes to determine how the genes segregated relative to one another (dihybrid cross)

-yellow/wrinkled crossed with green/round yielded an F1 that was yellow and round (heterozygous though)

-when this dually hetero yellow/round F1 was selfed, 16 possibilities exist

-use the product rule

**Punnett's square is an application of product rule

**Forked diagrams are an organized way of writing down the product rule, such that all possible outcomes at one event are multiplied by all possible outcomes of another independent simultaeously occurring event.

-phenotypic ratio of F2 is 9:3:3:1 (9yellowround/3greenround/3yellowwrinked/1greenwrinkled)

****(3:1)(3:1) > binomial factor =

(9:3:3:1)!!

*****this is where we get Mendel's second Law : during gamete formation, the segregation of alleles of one gene is independent of the segregation of the alleles of another gene

Test Cross of a Dihybrid Hetero

-YyRr /w yyrr

-would expect 1:1:1:1

Message:

If a dihybrid individual is test-crossed or selfed, a 1:1:1:1 or a 9:3:3:1 ratio respectively indicates that the TWO GENES ARE ASSORTING INDEPENDENTLY

How does the position of the genes within the genome afect the inheritane pattern?

Bateson and Punnet (early 1900s)

-tried to replicate Mendel with dihybrid, yet 9:3:3:1 did not appear... instead dominant alleles are together and recessive alleles are together

-P/L are coupled, p/l also

Thomas Morgan -Drosophila

-similar results with two Drosophila genes, one regulating wing development and the other eye colour

pr=purple eyes/pr+=red

vg=vestigal wings/vg+=normal wings

**experiment was a test cross, first with parental then with F1

-expect 1:1:1:1 in F2, yet again, largest classes are those in which wild type alleles are coupled or recessive alleles are coupled

HYP: The coupling of alleles is based on their arrangement in the parents, and not on their dominant/recessive relationships .

 -to determine if true, crossed individuals with dominant/recessive on same chromosome (ie. red with vestigal and purple with normal wings)

-this time, dominant allele pair is in repulsion

-***parental alleles are linked

Conclusion:The alleles which are together in the parents stay together through subsequent generations, independent of their dominane/recessive relationship

-these genes are on the same chromosome= linked

 cis-dyhibrid=dominant/recessive alleles are on the same side (alleles in coupling)

trans-dihybrid=dominant paired with recessive on the same chromosome (alleles in repulsion)

Y/y;R/r and Y/y.R/r

- slash denotes either side is one of a homologous pair

-semi-colon indicates two different chromosomes

-period indicates unkown relationship

1)Develop null

-have to have a reason

-have to have word 'significant'

-have to have the ratio

ie: The genes, A and B, are independently assorting (ie. not linked) and the test cross progeny do not deviate significantly from the expected 1:1:1:1 ratio.

2) Calculate X²

3)df

4)compare test statistic to critical value

-if it is less than critical, then p is greater than 0.05, and fail to reject

-if greater, p is less, and reject

5)evaluate null

Alfred Sturtevant (1911)

HYP: The frequency of recombinant products of meiosis is related to the distance b/w genes

-the further apart genes are, the more likely a cross-over iwll occur b/w them

-increased distance = increased recombinant meiotic products

-created a unit (map unit, later called a centiMorgan, cM) which was equivalent to the frequency of recombination:

...1 genetic map unit (mu) = 1% recombinant products of meiosis= 1 cM

Conclusion: Using teh recombination frequency b/w linked genes, one can determine the position of genes relative to one another

genetic locus= position of a generation

-recombinants are usually the small groups/classes

-map distance = percent recombinant progeny

= (# recombinant progeny/total prog.) *100

Note: The frequency of recombinant products never exceeds 50%, therefore the farthest distance that can be calculated b/w 2 genes is 50cM (max 1 crossover)

-use sum of smaller parts

-they will assort independently

-behave as if on different chromosomes

-No crossing over in male (XY)

-X-linked genes in Drosphilia = w+(allele for eye colour; red/white), f+(allele for bristle type; unforked or forked)

Experiment

-because the male is hemizygous, crossing to a mutant F1 fly has teh same effect as crossing to a tester (homozygous recessive fly)

-therefore, instead of testcross, carry through to the F2

If we know the map distance b/w two genes, we should be able to predict the frequency of recombinant and parental phenotypes resulting from a given cross.

If A and B assort independently, you get 25% of each allele (ie. 1:1:1:1) of F1 test cross, therefore 50% parental and 50% recombinant.

If genes C and D are linked, and we know they are 15cM apart, then we know that 15 % are recombinant classes.

Therefore, if F1  cross, get 15% of two types of recombinant (CD and cd) and 85% of parental types (Cd and cD)

Arabidopsis chromosome 5 contains these 3 genes (ie. they are linked):

ify+/ify - leaves

cer3+/cer - wax/no wax

yi+/yi - green/yellow stem

Test cross, using all 3, will have large number of parentals, as usual, lower number of recombinants(SCO), and VERY low number of DOC (double cross over)

Long Method

calculate distance b/w each set of points

-then determine orientation of genes

...there are crossover hotspots and deadzones... 

What is the probability of double crossovers?

p(DCO)= p(SCOI) + p(SCOII)

-this assumes that the cross-overs are independent of one another, that having a cross-over in one region does not inhibit/promote a cross-over in another region

interference= a measure of how much one cross-over has interfered with another

positive interference = fewer DCO than expected

negative= more DCO than expected

interference = (observed DCO)/(expected DCO)

=observedDCO/(pSCOI*pSCOII*1000)