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- 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
- 4. THE CELL CYCLE
- 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
- 6. MITOSIS VERSUS MEIOSIS
- 7. UNIQUELY MEIOTIC Leptotene, Zygotene, Pachytene, Diplotene, Diakinesis
- 8. EXERCISE 2 Genotypes Versus Phenotypes and other Important Terms Application of Mendelian Laws Probability Concepts
- 9. TERMS...TERMS...TERMS
- 10. GENOTYPES & PHENOTYPES
- 11. MENDELIAN INHERITANCE
- 12. MONOHYBRID CROSS
- 13. DIHYBRID CROSS
- 14. TRIHYBRID 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
- 22. ALLELIC INTERACTIONS
- 23. INCOMPLETE DOMINANCE 1:2:1 phenotypic ratio
- 24. CODOMINANCE 1:2:1 phenotypic ratio
- 25. DOMINANT LETHAL
- 26. RECESSIVE LETHAL
- 27. NON-ALLELIC INTERACTIONS
- 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

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