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POST-LAB
DISCUSSION 1
   Exercises 1 to 3




                      PARUNGAO 2010
EXERCISE 1
Demonstration 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 HOURS
 INTERPHASE (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 species
    G2 (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 2
Genotypes 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 EXPANSION
Mathematical 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
APPLICATIONS
1 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

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

                     B (agouti) and b (black)
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 EVENTS
Either 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 3
Analyze 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 2
                     If W is white and Y is
           WHITE       yellow


                     W is dominant to w
           WHITE
                     Y is dominant to y

           YELLOW W is epistatic to Y and y


                     Y is epistatic to ww
           WHITE
                     W and yy same expression
RECESSIVE EPISTASIS
         B_: 	 agouti
         bb:	   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 w
ww epistatic to P
P dominant to p
pp epistatic to W
The 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 PHENOTYPE
           A dominant to a
           B dominant to b


           A interacts with B
             producing new
             phenotype


           aabb produces
             fourth
             phenotype

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Bio 140 lab discussion 2010 ex 1 to 3

  • 1. POST-LAB DISCUSSION 1 Exercises 1 to 3 PARUNGAO 2010
  • 2. EXERCISE 1 Demonstration and Observation of Chromosomes = KNOW YOUR PROCEDURES! REVIEW! The Cell Cycle Mitosis Versus Meiosis
  • 3. Purpose of Carnoy’s Fluid = fixative Purpose of Giemsa = staining the chromosomes Onion root tip = active cell division
  • 5. TIMING = 24 HOURS INTERPHASE (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 species G2 (3 to 4 hours) MITOSIS (2 hours) the cell makes preparations for and completes cell division only takes about 2 hours
  • 7. UNIQUELY MEIOTIC Leptotene, Zygotene, Pachytene, Diplotene, Diakinesis
  • 8. EXERCISE 2 Genotypes Versus Phenotypes and other Important Terms Application of Mendelian Laws Probability Concepts
  • 12. MONOHYBRID CROSS
  • 15. BINOMIAL EXPANSION Mathematical 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
  • 16. APPLICATIONS 1 OFFSPRING (a + b) 2 OFFSPRING (a + b) 2 3 OFFSPRING (a + b) 3
  • 17. What if you are only after a certain combination? use of FACTORIAL P = [n!/x! (n-x)!] pxqn-x What is the probability is that a monohybrid cross yielding a litter of four pups will produce three agouti and one black pup? B (agouti) and b (black) 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
  • 18. MUTUALLY EXCLUSIVE EVENTS Either 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:
  • 19. 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
  • 20. EXERCISE 3 Analyze and Determine Gene Interaction Types
  • 21. ALLELIC VERSUS NON-ALLELIC ALLELIC only one gene controls one trait NON-ALLELIC two genes control one trait
  • 23. INCOMPLETE DOMINANCE 1:2:1 phenotypic ratio
  • 24. CODOMINANCE 1:2:1 phenotypic ratio
  • 28. 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)
  • 29. DOMINANT EPISTASIS: CASE 2 If W is white and Y is WHITE yellow W is dominant to w WHITE Y is dominant to y YELLOW W is epistatic to Y and y Y is epistatic to ww WHITE W and yy same expression
  • 30. RECESSIVE EPISTASIS B_: agouti bb: brown Presence of one C: allows pigmentation to occur Presence of cc: albino Ratio: 9:3:4 (example: mouse coat color)
  • 31. DUPLICATE RECESSIVE GENES W dominant to w ww epistatic to P P dominant to p pp epistatic to W The 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)
  • 32. DUPLICATE DOMINANT GENES two or more genes have the same effect on a given trait
  • 33. DUPLICATE GENES WITH CUMULATIVE EFFECTS Both gene pairs influence fruit shapes: cumulative
  • 34. NOVEL PHENOTYPE A dominant to a B dominant to b A interacts with B producing new phenotype aabb produces fourth phenotype