bio 140 lab discussion 2010 ex 1 to 3
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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
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
EXERCISE 2Genotypes Versus Phenotypes and other Important Terms
Application of Mendelian Laws
Probability Concepts
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
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
ALLELIC VERSUS NON-ALLELIC
ALLELIC
only one gene controls one trait
NON-ALLELIC
two genes control one trait
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)