Approach to the modern synthesis

Timeline of historical events in biology

http://www.biologie.uni-hamburg.de/b-online/e01/geschichte.htm

Timeline of historical events in genetics

http://las.alfred.edu/~emmonsc/Site/bio365/timeline.htm

Mendel’s dihybrid crosses

Mendel also did crosses where he kept track of two characters

Dihybrid (two-factor) crosses

Found that when he followed two traits, they were inherited independently

Crossed yellow/round x green/wrinkled

All progeny in F1 generation were yellow/round
Diagram:
GGWW x ggww = GgWw

Also crossed yellow/wrinkled x green/round

Again, all progeny in F1 were yellow/round
Diagram:
GGww x ggWW = GgWw

To produce F2 generation

Cross two individuals from the F1
Got this wild outcome

What creates this collection and ratio of phenotypes?

Mendel postulated that seed coat color and seed coat texture were inherited independently

So, used Product Rule to predict phenotype frequencies

Can do this in two ways

First method:  Predicting independent phenotype frequencies

Summary

Use Product Rule and Punnett square for each monohybrid cross to predict independent phenotype frequencies
Then use Product Rule to predict joint phenotype frequencies

Steps

First, use monohybrid cross for coat color or predict coat color phenotypes
Then, use Summation Rule to obtain phenotype frequencies

P(Yellow) = ¼ + ¼ + ¼ = ¾

P(Green) = ¼

Next, use monohybrid cross for coat texture to predict coat texture phenotypes
Then, use Summation Rule to obtain phenotype frequencies

P(Round) = ¼ + ¼ + ¼ = ¾

P(Wrinkled) = ¼

Then use Product Rule to predict joint phenotype frequencies
¾ yellow x ¾ round = 9/16 round yellow
¼ green x ¾ round = 3/16 round green
etc.
Here’s another way of illustrating this same process:
 

Second method:  Predicting joint gamete frequencies

Summary

Use Product Rule to predict joint gamete frequencies
Use Product Rule to predict phenotype frequencies

Steps

Use Product Rule to predict joint gamete frequencies

Possible gametes from RrYy
P(RY gamete) = ½ R x ½ Y = ¼
P(Ry gamete) = ½ R x ½ y = ¼
P(rY  gamete) = ½ r x ½ Y = ¼
P(ry gamete) = ½ r x ½ y = ¼

Use joint gamete frequencies to generate a large Punnet Square

Punnett square uses Product Rule to predict genotype frequencies
e.g. RY fuses with RY to produce RRYY
Therefore P(RRYY) = P(RY) x P(RY) = ¼ x ¼ = 1/16

Then use Summation Rule to predict phenotype frequencies
P(Round, yellow) = P(R-Y-) = P(RRYY) + P(RrYY) + P(RRYy) + P(RrYy) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 +1/16 + 1/16 = 9/16

Mendel’s Fourth Postulate

All these methods depend on Mendel’s Fourth Postulate

Mendel observed that each trait was inherited independently

Followed Product Law
Independent assortment of heritable units

Later, we’ll discuss situations in which traits are not strictly independently inherited

But, this is a good predictive model

Any observed mathematical deviations from the model point to situations that need more investigations
As we will see shortly

To test your understanding, try some dihybrid cross problems:

http://www.biology.arizona.edu/mendelian_genetics/problem_sets/dihybrid_cross/dihybrid_cross.html

Chromosomal Theory of Inheritance

1902-1903 Walter Sutton and Theodor Boveri

Noted parallelism between chromosome behavior and Mendelism

Precise sorting and recombination of chromosomes in formation of germ cells was striking
Independently proposed that each egg or sperm cell contains only one member of each chromosome pair

Mendel postulated that each individual has two copies of each heritable unit controlling a trait, one inherited from each parent

Likewise, Sutton and Boveri both observed that each individual has two copies of each type of chromosome

A gamete receives only one member of each chromosomal pair
Which would allow progeny to receive one heritable unit from each parent

1902 Walter S. Sutton (1877 - 1916)

While a graduate student in E. B. Wilson’s lab at Columbia University

Observed homologous pairs of chromosomes in grasshopper cells

These grasshopper chromosomes have quite distinct shapes, so can follow individual chromosomal pairs
Found that during meiosis the chromosome pairs split, and each chromosome goes to its own cell

The segregation pattern of chromosomes during meiosis matched the segregation patterns of Mendel’s genes.

He suggested that Mendel's "factors" must be located on chromosomes

"…the association of paternal and maternal chromosomes in pairs and their subsequent separation during the reducing division as indicated above may constitute the physical basis of the Mendelian law of heredity" (Sutton, 1902).
Sutton W.S. 1902. On the morphology of the chromosome group in Bracystola magna. Biol. Bull. 4:24-39.

In 1903, he made an even stronger argument to connect Mendel’s laws of heredity and the behavior of chromosomes in his paper: The Chromosomes in Heredity.

Sutton reiterated earlier work of Theodor Boveri, who in the late 1880s and early 1890s observed that chromosome numbers are cut in half as egg cells mature, and concluded that sperm and egg nuclei have half sets of chromosomes.

1902 Boveri

Discovered that a particular combination of chromosomes, rather than their number, is essential for normal cel development

Abnormal fertilization of sea urchin eggs by two sperm cells causes unequal distribution of chromosomes in resulting daughter cells

For example, one copy of chromosome c is attached to poles 1 and 2, and one copy is attached to poles 2 and 3
Thus, when chromosomes are segregated to the four poles at cell division, some daughter cells have too many copies of certain chromosomes and some have too few
For example, cell 2 has two copies of chromosome c and cell 4 has none.
Boveri found that these abnormal zygotes could not complete development

Today’s view of chromosomal inheritance

Single molecule of DNA in each sister chromatid

During meiosis and mitosis, packaged into four-armed structure

Each set of arms (a short and a long) is a chromatid

Mitosis

In mitosis, the pairs duplicate and one set goes to each progeny cell

Giving the daughter cells a complete complement of all the information in the parent cell

Mitotic division is used for reproduction by single-celled organisms

Hence, they are clones of one another
A lineage of bacteria are essentially genetically identical

Mitosis provides for cell proliferation during growth in a multicellular organism

Gametogenesis

However, in gametogenesis, the number of chromosomes must be reduced.  Why?

To avoid accumulation of chromosomes during fertilization

Meiosis

Gemete formation in dihybrid cross

Important points

First meiotic prophase
Every 4-armed chromosome pairs up with its homolog
One homolog was from mother, one was from father
First meiotic anaphase
Homologs segregate to one side or other of cell
Non-homologous chromosomes assort independently

In some pairs, father’s copy goes to right

In some other pairs, mother’s copy goes to right

Second meiotic anaphase
Chromatids separate

Every gamete gets a complete set (one member of each pair)

So, at fertilization, every chromosome will have a homolog (back to two units)

Chromosomal sex-determination

1905 Edmund Beecher Wilson (1856-1939) and Nettie Maria Stevens (Westford,USA, 1861-1912) (portrait from 1904)

Studying insects, independently propose that separate X and Y chromosomes determine sex.

Stevens, studying the scarab Tenebrio molitor shows that a single Y chromosome determines maleness, and two copies of the X chromosome determine femaleness.

Her work on sex determination was published as a Carnegie Institute report in 1905. She died in 1912 of breast cancer.

Biographical info:  http://www.dnaftb.org/dnaftb/concept_9/con9bio.html

In many organisms, sexes are chromosomally determined

Homogametic sex  -- that sex containing two like sex chromosomes

In most animals species these are females (XX)
Butterflies and Birds, ZZ males

Heterogametic sex --- that sex containing two different sex chromosomes

In most animal species these are XY males
Butterflies and birds, ZW females
Grasshopers have XO males

Sex determination in mammals

Patterns of sex linkage:

(a) Male, female progeny (sometimes) differ.

(b) Reciprocal crosses differ.

(c) For genes on the X chromosome, recessive alleles "appear" more often in males.

Punnett was a poultry geneticist

During the First World War, Punnett developed a technique of separating male and female chicks using sex-linked plumage colours.

In this way, the less useful male chicks could be separated from the more useful female chicks and destroyed.
This improved the efficiency of the poultry industry.

The first autosexing breed developed was the Cambar back in 1929 (Gold Campine X Barred Rock).

Barred plumage (alternating black and white bands on feathers) is a dominant sex-linked trait in chickens.

Remember, in birds, males are heterogametic

But, because barred is dominant, a male needs only one copy of the barred allele for its plumage to be barred
Punnett developed a breeding strategy using sex-linked plumage color to differentiate between newly-hatched male and female chicks
In Punnett’s strategy, the female parent must carry the dominant allele for the sex-linked trait
In this case, all male progeny get a recessive non-barred allele from their father, but the dominant barred allele from their mother makes them barred
Female progeny get only the non-barred allele from their father, and so are non-barred
Pictures illustrate the Cream Legbar breed
Leghorn x Barred Rock
As an exercise:
Try doing a Punnet square for this cross
Then, do a Punnet square for the reciprocal cross (where mother is non-barred and father is barred)

Homogametic sex is mosaic

Dosage compensation – shut down one copy of long arm

Random from one cell lineage to another
 
Creates patchiness – mosaic

e.g. in cats, orange locus on sex chromosome

Females are heterogametic Oo
Leading to a tortoiseshell cat
Only way to get completely orange cat is for it to be male: O-
For more details, see handout on cat coat-color genetics (Rich make a link here)
 
 

Gene interaction (epistasis)

1905 William Bateson and Reginald Crundall Punnett (1875–1967)

Found various deviations from the normal dihybrid ratio (9:3:3:1), which they rightly attributed to gene interaction.

They analyzed the three comb types of chicken known to exist at that time:

Chicken Varieties

Phenotype

Wyandotte

Rose Comb

Brahmas

Pea Comb

Leghorns

Single Comb

Result

The F1 differed from both parents
New, walnut phenotype
Two new phenotypes not seen in the parents appeared in the F2

Analysis

How can this result be explained?
Work this out at your desk

Talk to your neighbors

Clues:
The first clue is the F2 ratio (9:3:3:1)

Same ratio as F1 from a sib crossing of dihybrid cross progeny

This observation suggests that two genes may control the phenotype of the comb

Make a dihybrid cross Punnett Square

A series of experiments demonstrated that the genotypes controlling the various comb phenotypes were as follows.

Phenotypes

Genotypes

Frequency

Walnut

R_P_

9/16

Rose

R_pp

3/16

Pea

rrP_

3/16

Single

rrpp

1/16

What test crosses are needed to determine parental phenotypes?

The gene interactions and genotypes were determined by performing the appropriate testcrosses

Two possible genotypes for each parent

Rose:  RRpp

Pea:  rrPP

Or

Rose Rrpp

Pea:  rrPp

It was later shown that the genotypes of the initial parents were:

Rose = RRpp

Pea = rrPP

Therefore, genotypically the cross was:

 

Walnut

R_P_

9/16

Rose

R_pp

3/16

Pea

rrP_

3/16

Single

rrpp

1/16

This observation suggests that two genes may control a single phenotype:  comb morphology.

Epistasis

The interaction between two or more genes to control a single phenotype

Summary

The interactions of the two genes which control comb type was revealed because we could identify and recognize the 9:3:3:1. Other genetic interactions were identified because the results of crossing two dihybrids produced a modified Mendelian ratio. All of the results are modifications of the 9:3:3:1 ratio.

Things to think about:

Discuss Bateson’s and Punnett’s results from their crosses to examine the genetics of chicken comb morphology

How do their results differ from Mendel’s classic dihybrid crosses on wrinkled vs. smooth and green vs. yellow peas?
What can you say from this comparison about the hierarchical nature of the phenotype?