POST-LAB DISCUSSION 1

Exercises 1 to 3

PARUNGAO 2010

EXERCISE 1Demonstration and Observation of Chromosomes = KNOW YOUR PROCEDURES!

REVIEW!

The Cell Cycle

Mitosis Versus Meiosis

Purpose of Carnoy’s Fluid = fixative

Purpose of Giemsa = staining the chromosomes

Onion root tip = active cell division

THE CELL CYCLE

TIMING = 24 HOURSINTERPHASE (18-20 hours)

G1 (10 hours) is typically the longest phase of the cell cycle since it follows cell division in mitosis; first chance for new cells have to grow. S (5 to 6 hours) phase varies according to the total DNA that the particular cell contains which is fairly constant between cells and speciesG2 (3 to 4 hours)

MITOSIS (2 hours)the cell makes preparations for and completes cell division only takes about 2 hours

MITOSIS VERSUS MEIOSIS

UNIQUELY MEIOTIC

Leptotene, Zygotene, Pachytene, Diplotene, Diakinesis

EXERCISE 2Genotypes Versus Phenotypes and other Important Terms

Application of Mendelian Laws

Probability Concepts

TERMS...TERMS...TERMS

GENOTYPES & PHENOTYPES

MENDELIAN INHERITANCE

MONOHYBRID CROSS

DIHYBRID CROSS

TRIHYBRID CROSS

BINOMIAL EXPANSIONMathematical way to determine or project combinations

Let a = probability of first event; b = probability of the alternative event; and a + b = 1

THUS...a 50% probability or chance that one character will appear over the other

APPLICATIONS1 OFFSPRING

(a + b)

2 OFFSPRING

(a + b)2

3 OFFSPRING

(a + b)3

What if you are only after a certain combination?

use of FACTORIAL

P = [n!/x! (n-x)!] pxqn-x

B (agouti) and b (black)

What is the probability is that a monohybrid cross yielding a litter of four pups will produce three agouti and one black pup?

P =

n = # trials (births) (4)s = agouti (p = 3/4 = .75)

t = black (q = 1/4 = .25)

Therefore,P = [4!/3!1!](.75)3(.25)1 = 0.42

MUTUALLY EXCLUSIVE EVENTSEither one or the other will occur

focus is on the concept of outcome A or B

Example: In rolling a dice: calculate the probability of either two 4s or two 5s

Because these outcomes are mutually exclusive, the sum rule can be used to tell us that the answer is 1/36 + 1/36 which is 1/18. This probability can be written as follows:

What if: chances of both?The product rule states that the probability of independent events occurring together is the product of the probabilities of the individual events.

consider two dice and calculate the probability of rolling a pair of 4s

The probability of a 4 on one die is 1/6 because the die has six sides and only one side carries the 4

Therefore, with the use of the product rule, the probability of a 4 appearing on both dice is 1/6 × 1/6 = 1/36

EXERCISE 3Analyze and Determine Gene Interaction Types

ALLELIC VERSUS NON-ALLELIC

ALLELIC

only one gene controls one trait

NON-ALLELIC

two genes control one trait

ALLELIC INTERACTIONS

INCOMPLETE DOMINANCE

1:2:1 phenotypic

ratio

CODOMINANCE

1:2:1 phenotypic ratio

DOMINANT LETHAL

RECESSIVE LETHAL

NON-ALLELIC INTERACTIONS

DOMINANT EPISTASIS: CASE 1

W is dominant to w (W white) Y is dominant to y (Y yellow) W is epistatic to Y and y

In the absence of a dominant allele, the YY or Yy yellow while yy another phenotype which is green)

DOMINANT EPISTASIS: CASE 2If W is white and Y is

yellow

W is dominant to w

Y is dominant to y

W is epistatic to Y and y

Y is epistatic to ww

W and yy same expression

WHITE

WHITE

WHITE

YELLOW

RECESSIVE EPISTASISB_: agoutibb: brown

Presence of one C:allows pigmentation to occur

Presence of cc: albino

Ratio: 9:3:4 (example: mouse coat color)

DUPLICATE RECESSIVE GENES

W dominant to www epistatic to PP dominant to ppp epistatic to WThe presence of at least one dominant allele of each two gene pairs is essential for flower to be purple

Ratio: 9:7 (Sweet Pea flower color)

DUPLICATE DOMINANT GENES

two or more genes have the same effect on a given trait

DUPLICATE GENES WITH CUMULATIVE EFFECTS

Both gene pairs influence fruit shapes: cumulative

NOVEL PHENOTYPEA dominant to aB dominant to b

A interacts with B producing new phenotype

aabb produces fourth phenotype