Genetics: 08

Wednesday

Genetics Lecture 14, 10/1: Extranuclear inheritance, Sexual determination in Drosophila

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TEST 2 MATERIAL

• We are responsible for epistasis problems for the test (started it before exam 1)
• Sex-linked traits
• Nature vs. nurture. Multiple choice, short answer, essay
• 3 to 4 problems comprising 50% of the credit. Mult. Choice is 20% and short ans. 30%.
• Up through organelle heredity
• Review is at NOON

• Sex determination in Drosophila (unique system where they are more than male or female) (6:25)

o 1916 Calvin Bridges

• looking for the sex determination in fruit flies. He found that the mechanism was not Y dependent. He found that it was the ratio of X chromosome to autosomes determined the sex of the fly.

o normal diploid female has 2 autosomes for every 2 x chromosomes. Ratio of 1.0
o Triploid female extra copy of autosome and x chromosomes: total 3 autosomes to 3x chromosomes - ratio of 1.0

o any arrangement with 1.0 would be a female
o 3 x chromosomes for 2 autosomes gives ratio of 1.5. These are called metafemales: weak, infertile, low viability
o Males have ratio of 0.5 two cases.

• one x chromosome to two autosomes. Lack the Y chromosome leaving them sterile
• One x chromosome one y chromosom --> fertile. Y chromosome determinse fertility, not maleness.

o Metamales (14:20)

• Ratio of 0.33. 1 x chromosome fore every 3 autosomes. They are weak and infertile

o 2 classes of flies with ratios between 1 and .5 - Intersex flies

• Intersex flies come about when there is: 3 X Chromosomes to 4 autosomes or 2 X chromosomes to 3 autosomes
• Intersex flies: rudimentary gonads, sterile morphological, display both male and female (end of chap 7)

• Chapter 9 (17:45) Extranuclear inheritance: anytime inheritance is not base on the DNA in the cells nucleus.

o Ex. Organelle heredity – have DNA in the chloroplast and the mitochondria which influence phenotype. Always mother to offspring – egg contributes all of the cytoplasm → Organelles are in the cytoplasm.
o Ex. Infectios Heredity – phenotype associated with a microorganism living in a cell.
o Ex. Maternal effect – proteins in the egg which determine a particular phenotype in the offspring.
o Only organelle heredity is DNA based

• ORGANELLE HEREDITY (24:00)

o 1908 Carl Correns worked with four O’clock plants. Either have white leaves, green, or veriegated leaves.
o If the progeny came from a fertilization of an ovum from a plant with green leaves then the progeny had all green leaves.
o Phenotype of the offspring was always the same as the phenotype of the parental plant that gave the ovum.
o Phenotype is associated with the DNA of the chloroplast.

Mitochondria can show the same pattern (28:10)

o 1952 Mary and Hershey Mitchell studying Neurospora (bread mold) cross. Discovered a strain that they named Poky Strain (it grew slowly). Determined that the phenotype was associated with a mutation in the mitochondrial DNA. The mitochondria generates energy through respiration. Thus a mutation in the mitochondria would result in the slow growth phenotype.

o Female poky x male WT THEN all progeny are were poky

o female WT x male poky THEN all progeny were WT

o The question: Do chloroplasts and mitochondira have their own DNA? → Yes
o Cytoplasmic inheritance occurs through the cytoplasm of the egg. The sperm contributes NO cytoplasm.
(35:30) DNA of the chloroplast

o Uniform in size and shape and ranges from 100kB to 225kB (kb=kilobases)
o Circular double stranded DNA with no associated proteins.
o Genes found in chloroplasts DNA – genes for rRNA, tRNA ribosomal proteins and RNAPOL. This allows them to do their own transcription and translation independent o the cells individiual mechanisms for transcription and translation. They also have genes for photosynthesis.

(38:40) Mitochondrial DNA

o 16kB to 36kB
o rRNA’s 22 different tRNA’s
o genes whose products play a role in oxidative phosphorilation.
o Ribosomes of mitochondria are quite different than the ribosomes of the rest of the cell.

• Mitochondrial mutations associated with human diseases (41:30)

o DNA in the mitochondria is particularly likely to be mutated
o How do we determine that a human disease is mitochondrial in nature?

• Inheritance pattern should be maternal and not mendelian.
• Deficiency must result based on loss of energy for the cell
• You must have a mutation in the gene in the mitochondrial DNA associated with the disease.

o Myoclonic epilepsy and ragged red fiber disease (45:25): lack of coordination, deafness, dementia, epileptic seizures mutation in the mitochondrial gene for tRNA
o Lebers hereditary Optic Neuropathy: Sudden bilateral blindness, avg. onset of 27
o Kearns Sayre Syndrome: Hearing loss, loss of vision and heart condition. Normal childhood with an onset in early adulthood.

This will be material for TEST 3 ----------

• INFECTIOS HEREDITY (50:25)

o Eukaryote living in a symbiotic relationship with a microorganisms and leads to a phenotype.
o EX. Paramecium Aurelia → a toxic substance which kills sensitive cells. Maintained by the Kappa gene.
o Dominant K (kappa particle) allele to have the killer phenotype. You also need the microorganism which requires cytoplasmic exchange during paramecium mating.
o Conjugation with cytoplasmic exchange or without.

Monday

Genetics Lecture 13, 9/29: Primary Sexual determination, Genetic diseases, dosage compensation

• Human Embryonic, sexual differentiation (primary)

o A relatively early process.
o By the 5th week of gestation developmental gonads have appeared. Initially associated with the kidney.

• Outer cortex

Can become an ovary

• Inner mendula

Has the ability to develop into testes.

o Ducts

• Male → wolffian ducts
• Female → mullarian ducts
• The SRY has the testes determining factor. If this is present the development of the testes and male ducts is triggered. If it is NOT present (no Y chromosome) by week 12 we start to produce oogonia. Then we begin meiosis and by the end of the 25th week meiosis I is complete and all of the potential eggs are present in the female.

• What can go wrong with sex chromosomes. (7:25)

o Non-disjunction event: during meiosis, the sex chromosome pair fails to separate so you get unequal distribution of sex chromosomes in the gametes.
o This leads to a number of syndromes associated with unusual sex chromosome arrangements.
o Kleinfelter syndrome: males with two X chromosomes and a Y chromosome. These individuals are labeled 47 XXY where 47 is the number of chromosomes and XXY is the arrangement of the sex chromosomes.

• Individuals can also have other arrangements: 48 XXXY, 48 IIYY, 49 XXXXY, 49 XXXYY
• Generally the greater the number of X chromosomes the worse the syndrome.
• Normal male ducts and normal genitals but the testes fail to form properly. The result is sterility.
• Unusually long arms and legs, large hands and feet, undergo some female secondary sexual development including rounding of hips and some breast growth, below avg. intelligence.

o Turner syndrome (14:40)

• Also known as “45X”. Missing an X chromosome.
• Female with normal female genitalia, normal female ducts, incomplete ovaries → sterility
• Short in stature, skin flap on back of their necks

o 47 XXX syndrome

• some have it and are normal
• others are underdeveloped sex characteristics, below avg. intelligence and sterile
• ALSO: 48 XXXX and 49 XXXXX → traits become more pronounced.

o 47 XYY syndrome (20:00)

• Male, above average height. Below avg. intelligence, and socially awkward.
• Theorized that this is more common in a prison population then the population at whole (controversial).

• Ratio of males to females (not always one:one)

o in theory, a heteromorphic (male) individual should make equal number of X and Y gamets.
o Look at two ratios of males to females in the population
o (25:25) Primary sex ratio: Ratio of male to female conceptions
o Secondary Sex ratio: male to female births
o Secondary sex ratios:

• Caucasians: 1.06 (106 males to 100 females)

o Question: Does the primary sex ratio 1.6 mean that more females die in the embryonic period than males?

• observe miscarriages and abortion
• primary sex ratio can be anywhere from 1.08 to 1.6 → this suggests that there are more male conceptions than female.

o Primary sex ratio of 1 assumes:

• Male will produce same number of x and y gamets
• Each gamete type has the same viability in the reproductive tract
• The egg is equally receptive to both gamete types.

o (32:45) Current theory as to why “more males are born than females” is that the gamete bearing the Y chromosome is lighter and therefore faster than the gamete with an X chromosome.

• Dosage compensation (35:30)

o All of the autosomal chromosomes come in pairs.
o Sex chromosomes – females have 2 X chromosomes males have 1 X chromosome. Twice as many copies of the X chromosome genes in females
o Problematic – need to balance the amount of genes in all sexes.
o Barr and Bertman – looking at DNA in interphase and they identified a dark staining body in the nucleus of females. They called it a Barr body. In females, one of the two X chromosomes is inactivated.
o Barr body: inactivated X chromosome
o How is the inactivation of one X chromosome determined?

• Lyon hypothesis – the inactivation of the X chromosome is random. The same x chromosome is inactivated in every cell of an individual

o Work to support lyon looks at:

• Glucose – 6 – phosphate dehydrogenase (X chromosome gene)
• Looked at a large number of individuals that were heterozygotes for the gene.
• If the lyon hypothesis is correct then 50% of these will have the wildtype copy of the gene and 50% will have the mutant. (they did the test and it is correct – accepted as true)

o How do you make a Barr Body and thus inactivate the X chromosome? (46:50)

• A region on the X chromosome is responsible for this called the X inactivation center.
• X inactivation center contains 4 genes: one of these genes when activated produces a functional RNA. It is Believed that this RNA covers the X chromosome that is inactivated. The coating is what prevents the X chromosome from functioning.

Sunday

Genetics Lecture 12, 9/26: Sex determination and Sex Chromosomes

• The chromosomes that determine sex are different.
• Heteromorphic chromosome pair – a chromsome pair which is not homologous.

o Ex. XY chromsome pair in human males

• Multicellular organisms go through a two step sexual differentiation process.

o Primary sexual differentiation. Embryonic, intial determination of male or female
o Secondary sexual differentiation (puberty).

• In general terms, organisms have one of two systems for sexes.

o Unisexual organisms – the organism has one type of gonad, male or female
o Bisexual – one organism with both gonads AND are fertile/functional

• Life Cycles

o Chlamydomonas – organism that rarely undergoes sexual reproduction. Spends the majority of the lifecycle as a haploid organism. Reproduce here through asexual reproduction. When the cells are stressed → nitrogen depletion. They will form isogametes, they will have two mating types, + and - . Mating is mating type dependent, plus with minus and vice versa. In turn you produce a diploid zygote. This diploid zygote will help protect the cell from the harsh conditions. When conditions improve the zygote undergoes meiosis and you produce vegetative cells (haploid).

• C. Elegans (17ish)

o Two sexes, hermaphrodites or males. Hermaphrodites have testes and ovaries and can make both functional gametes therefore they can self fertilize
o 99% of worms are hermaphrodites 1% are males
o hermaphrodites self fertilize – 99% herm. And 1% male
o mate a male with a hermaphrodite you get 50% male and 50% herm.
o In C. Elegans you have no Y chromosome so a hermaphrodite has two X chromosomes and the male has one X chromosome. SO when you self fertilize the hermaphrodite. 1% of males, due to non disjunction. Hermaphrodite produce a sperm or egg with no sex chromosomes.
o Cross male and hermaphrodite with no sex chromosomes you get half herm. And half male.

• (20ish)X and Y chromosomes were first linked to sex determination in the early 20th century. This was done in 1906 by Edmund Wilson.

o Observed an insect called the protenor. Females with 12 autosomal chromomse and 2 X chromosomes. Produce gametes with 6 autosomes and 1 x chromosome.
o Males have 12 autosomes and a sincle x chromosome. Produce two types of gametes. The first has 6 autosomes and x chromsome AND the other has 6 autosomes
o In the protenor, maleness is defined by the lack of chromosome.

• (24:30) – lygaeus mode of inheritance

o females have 12 autromse and 2 X chromosomes
o males have 12 autosome and Xand Y chromosome. Maleness is defined by the presence of a Y chromosome.
o Typically the male of the species is the heteromorphic organism. There are some organisms where the female is heteromorphic.

• Fish and reptiles – males are ZZ. Female is ZW
• Y chromosome is responsible for maleness in humans (lygaeus mode of inheritance)

o In 1920’s the Y chromosome was first visualized – small and heterochromatic.
o Originally thought that humans had 48 chromosomes (we have 46). 44 are autosomes and an XX chromosomes and XY for males.

• Y chromosome is genetically inert – very few important genes. HOWEVER – there are a number of important regions on the Y chromosomes.

o PAR – pseudo autosomal region: regions that share homology with portions of the X chromosome. Allow the X and Y chromsome to bond to each orther during mitosis and meiosis. Represents 5% of the Y chromosome. The remaining 95% is called the NRY
o NRY – non recombining region: this is genetically different from the X chromosome. 95% of the Y chromsome.
o SRY – sex determining region of the Y chromosome: within it is a gene for the testes determining factor. This is a gene that is believed to be responsible for the development of testes.

Genetics Lecture 11, 9/24: Linkage Mapping

• Three criteria for a successful mapping experiment

o Genotype of the organism which is producing crossover gametes needs to be heterozygous
o The cross should be constructed in a way to allow you to determine the genotype of the crossover chromosome by examining the phenotype of the offspring

• Easy way to do this , cross by homozygous recessive

o A sufficient number of offspring is necessary to allow you to understand the patterns.

• Three traits in drosophila

o Yellow body: y, white eyes:w, eye shape: e
o All three traits on the X chromosome
o final numbers
o 80 yellow body, WT eye color, WT eye shape (SCO 2)
o 193 yellow body, white eyes, WT shape (SCO 1)
o 3 yellow body, WT eyes, echinus
o 4685 all 3 mutant traits (DCO)
o 4759 all 3 WT traits (DCO)
o 3 WT body, white eyes, WT eye shape
o 207 wild type body, WT eye color, WT echinus (SCO 1)
o 70 WT body , white eyes, echinus (SCO2)
o You should be able to tell the relative order of the genes and the mapping distance between the genes.

• In any three factor cross data set there are 4 categories of data.

o The first is the non-crossover offspring. Offspring that inherit intact parental chromosome. Represent your largest class of offspring.

• In the drosphila it is the all WT and all Mutant, labeled NCO (non-cross over)

o Double crossover offspring (DCO). Offspring in which 2 crossovers have occurred. Smallest class in your population, least likely event.
o Single crossover 1: result of a single crossover between one of your gene pairs (relatively similar numbers)
o Single crossover 2: result of a single crossover between the 2nd gene pairs (relatively similar numbers)

• NCO: 4685 ywe, 4759 y+w+e+
• SCO1: 193 ywe+, 207 y+w+ e
• SCO2: 80 yw+e+, 70 y+we
• DCO 3 yw+e, 3y+we+
• Next step – determine the order of genes

o What genes are on which chromosome?

• Determine which genes are on which chromosome by examining the NCO

o What are our possible arrangements? NOTE: 3 factor crossover mapping will only tell you which gene is in the middle

• 3 possible alignments

 ywe
 wye
 wey

o perform theoretical double crossobers on all 3 of our possible alignments. Determine which one matches our actual DCO
o Our order is y w e , y+w+e+
o Determine the mapping distances between the genes
o identify all of the offspring with a cross between two genes, total number of offspring.
o Always need to add the DCO offspring to your number, these have a cross between y and w.

• Coefficient of confidence

o C=observed DCO/expected DCO = .0006

• Expected DCO is your mapping distances divdied by 100
• (1.56/100)*(4.06/100)
• DCOexp=.00063
• Interference is 1-C
• I = 1-.95 = .05

o Two types of interference

• Positive: fewer D-crossover than expected
• Negative: more D-crossover than expected

• CORN 42:50

o Bn – brown all WT have “+”
o Pr – purple
o V – virescent seedling
o F2 offspring

• 82 pr+ r+ bn SCO1
• 44 pr v n DCO
• 230 v bn pr+ NCO
• 42 pr+ v+ bn+ DCO
• 79 pr v bn+ SCO1
• 237 pr v+ bn+ NCO
• 200 pr+ v bn+ SCO2
• 195 pr v+ bn SCO2

o determine the order of traits

• 3 possible arrangements
• take the NCO and arrange them in 3 possible arrangements. Perform the double crosses and determine if it lines up with the DCO that is given. After doing this out the correct order is bn pr+ r, bn+ pr v+

o determine the distance

• perform a single cross on the above order and you get
• bn pr v+ (195), bn+ pr+ v (200) then add the DCO numbers (86) to get 481. Divide that by total number (1109) and you get 43.4. This is the distance from bn → pr+
• do this again for the distance from pr to r. bn pr+ v+ (82), bn+ pr v (79). 161 plus 86 gives you 247/1108. This gives you a distance of 22.3 mu between pr+ and r

o determine interference

• C=observed DCO/expected DCO
• .078/.097
• .434 * .223 = .097
• C= 0.8
• 1-0.8 = 0.2, a postive interference

Genetics Lecture 10, 9/22: Linkage Mapping in Eukaryotes

• Genes are found on chromosomes. There are times when genes found on a chromosome are transmitted in tandem. When this happens it impacts how traits are inherited.
• Meiosis – during prophase 1 homologous chromosomes pair together and crossing over occurs. There is exchange between the Homologous chromosomes. Frequency with which crossing over occurs can give us a relative distance between genes. The higher the frequency of crossing over, the greater the distance between genes. Put another way the closer two genes are, the more likely they will be transmitted together on a chromosome.

o The benefit of this is that it allows us to map genes on chromosomes. We are looking at the order of genes and the relative distance of genes.

• (10:10)
• ABC Mom and abc Dad
• Prophase I → 4 copies of each chromosome. Two types of chromosomes where crossing over occurs. The other two chromosomes will be the non crossover chromosomes. This will always be the highest % of chromosomes.
• 1 A B C
• 2 A B C
• 3 a b c
• 4 a b c
• 2 and 4 cross
• You produce a variety of crossover chromosomes based on where the crossover occurred. Within the crossover chromosomes the higher the % of two genes being transmitted together the closer those genes are.
• (17:05) Complete linkage altering phenotype. There are times when two genes are so closely linked that they are always transmitted together.
• If you cross heterozygotes with these traits the appearance of the F2 is altered.
• Mutant: Brown eyes + heavy wing vein
• Wildtype: red eyes + thin wing veins
• Parent: Brown eyes and thin wings (homozygous) X with red eyes and heavy wings. These are transmitted together and your F1 generation will have Brown Eye and thin wing as well as red eye and heavy wing. Phenotype red-eyes thin wing veins.
• Cross the F1 brown eye, thin wing + red eye heavy wing X brown eye, thin wing +red eye heavy wing

o Red eye =R
o Brown eye = r
o Thin wing veins = T
o Heavy wing veins = t
o Cross the F1 and you will have

• 1 brown eyes, thin wings
• 1 red eyes thin wings
• 1 red eyes thin wings
• 1 red eyes heavy wings.
• 1:2:1 final ratio

• Crossing over serves as the basis for mapping genes on a chromosome.
• 1911 – alferd sturterant and Thomas morgan.

o They were working with drosophila. Morgan crossed two flies. They were yellow bodied white eyed females X wildtype males
o Both of these traits are on the X chromosome.
o He produced and F1 which he crossed again to produce an F2. He found that 98.7% of his offspring had a parental phenotype. 1.3% had yellow body wtype eyes or WT body and white eyes.
o These were strange findings and they performed another cross. White eyed miniaterue winged females and WT males (both traits on X chromosome)
o This time 62.8% had a parental phenotype and 37.2% were wildtype eyes miniature wings or white eyes and wt wings.
o Morgan had two questions.

• What was the source of gene separation?

 ANS: came from cytological evidence. They were looking at chromosomes during meiosis. They saw that the homologous chromosomes wrapped around each other at the chiasmata. Morgan proposed that this served as the mechanism of seperation.

• Why did the frequency of seperation vary depending on the genes being studied.

 ANS: (41:25) Morgan proposed that the relative distance between genes determined the liklihood a chiasmata would form between two genes.
 Therefore the frequency with which genes crossover that share a chromosome is relative to the distance between them.

o Sturtaveant realized that this technique would allow one to map the position and relative distance of genes on a chromosome. Yellow body had a frequency of crossover to white eys of 0.5%. White eyes showed a frequency of crossover to miniature of 34.5%. Finally, yellow body showed a frequency of crossover of 35.4% to miniature.
o Sturtavant proposed that 1% of crossover would equal 1MU (mapping unit)
o 2 Additional observations by sturtavant

• crossing over occurs on both X chromosomes and autosomal chromosomes. In drosophila, crossing over only occurs in females.

Genetics Lecture 9, 9/19: Sex linked traits

I was not present during this lecture - my sister was getting married!

Genetics Lecture 8, 9/17: Epistasis and Sex linked traits

• (13:00ish) When we do epistasis problems we are going to work with a set of assumptions.

• 1st assumption: All phenotype classes will be unique from each other. This allows you to clearly identify one from the other.
• 2nd assumption: Assume that our traits are not linked and assort independently of each other. For simplicity we will label our alleles A, B, C.
• 3rd assumption: When complete dominance occurs we will designate AA and Aa as A_
• 4th assumption: Our crosses will be parental AABBxaabb. Therefore our F1 crosses are AaBbxAaBb
• 5th assumption: We will focus our analysis on the F2 generation. How do our relationships between the genes yield the final F2 ratio.

• Mice. Agouti (gray) is dominant A_ and black is recessive aa. Second locus in which the presence of the dominant allele gives coat color. B_ → gives coat color, bb → albino.

• Based on our assumptions . . .

o 9/16 A_B_ → Agouti
o 3/16 A_bb → Albino
o 3/16 aaB_ → Black
o 1/16 aabb → albino
o Final ratio is 9 agouti, 4 albino and 3 black.

• A dominant allele at one locus masks the genotypes at the 2nd locus
• EX. Fruit color in summer squash

o A dominant A allele = white fruit color
o If the A allele is recessive = aaB_ Yellow and aabb green
o 9/16 A_B_ → 9 white
o 3/16 A_bb → 3 white
o 3/16 aaB_ → 3 yellow
o 1/16 aabb → 1 green
o final ratio is 12:3:1

• EX. Fruit color in sweat peas. (27:45)

o Cross two white pea plants and produce 9/16 purple and 7/16 white. Proposed that pea color is controlled by two genes. Proposed two parent plants: AAbbxaaBB

• 9/16 A_B_ → purple .. 9/16
• 3/16 A_bb → white
• 3/16 aaB_ → white
• 1/16 aabb → white .. 7/16
• Propose that one dominant allele for each gene is necessary for purple color.

• Sex Linked Traits (34:10)

o Chromosomes → XX in female and XY in male. Autosomal chromosomes are everything else.
o In males the X and Y chromosomes synapse. But the Y chromosome has very little genetic information. The male phenotype is a direct reflection of the X chromosome from Mom. For a female there are two copies of the X chromosome and as a result of more normal dominant/recessive relationships. The end result is that you have unique patterns of expression for these traits.
o EX. (39:00) Eye Color in fruit flies

• Wildtype eye color is red and mutant color is white. Gene for this trait is on the X chromosome.
• Cross between wild type female x white eyed male. XwXw ¬x XwY
• Females are all XwXw
• Males are all XwY
• All offspring in the F1 gen are wildtype
• Cross the F1

 XWXwx XWY
 Females are all WT
 Males ½ Wildtype and ½ white-eyed

o One hallmark of a sexlinked trait is that there are different ratios for different sexes.
o Another hallmark of an X-linked trait is that reversing the parental phenotypes alters the ratio.
o White eyed female x red eyed male
o XwXw ¬x XWY → cross it to get the F1

• Females are all wildtype and males are all white
• F2 XWXw ¬x XwY →
• ½ females are red eyes.
• ½ xwxw white eyes
• Males ½ xWY redyes
• 1/2XWY white eyes

Genetics Lecture 7, 9/12: Review for Test 1

This is the audio for today's lecture. Whenever there is a time in parenthesis it refers to the time in the recording that he discussed that particular subject. It is a review of what will be on the 1st test. My notes from the review are as follows:

- No X linked or pedigree problems!!
- 4th post. → Independent assortment is at the level of the gene. Different genes – hair color eye color. Assort to gametes independently
- 3rd post→ different alleles of the same gene. We all have 2 alleles for a gene and there is an equal chance of any gamete getting each allele. (11:45)
- prophase 1 (5 phases)

MENDELS POSTULATES
1. Genetic characteristics are controlled by unit factors (genes) which exist in pairs in individuals. Two copies of each gene one from mom and one from dad.
2. When you have different forms of the same unit factor (alleles) in an individual one of the traits dominates over the other. Referred to as the dominant trait. In Mendels experiments it appeared solely in the F1 generation and 3/4 in the F2 generation. Recessive trait which appears as 1/4 of the F2. (this is true for a lot of situations but it is not universally true.)
3. During gamete formation the unit factors (genes) segregate independent of each other. Roughly an equal likelihood of a gamete receiving a particular trait. Basically if an individually has both genes (an allele for tall and one for short) 50% of gametes will be tall and the other half will be dwarf.
4. during gamete formation segregating pairs of unit factors (genes) assort independently of each other. Law of independent assortment.

Genetics