Lab 8: Population Genetics and Evolution Lab 9: Transpiration

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  1. When doing Hardy-Weinberg equilibrium, which value must you find first and why?
    q because p is both homo and heterozygous. So you'd have to find recessive first
  2. When looking for number of A alleles, what do you do?
    a alleles?
    • Multiply Homozygous Dominant individuals x2 + Heterozygous
    • e.g. 8 x 2+18= 34
    • Multiply Homozygous Recessive individuals x 2 + Heterozygous
    • e.g. 4 x 2 + 18= 26
  3. What does Hardy-Weinberg equation predict for the new p and q?
    • For a population, the equation states that the population would be at equilibrium if it remains at those particular frequencies.
    • For that population, it states that the value isn't equal as it was. Also, the new p and q will be determined by chance.
  4. In a large population, would it be possible to completely eliminate a deleterious recessive allele? Explain?
    no because the recessive allele is carried by the heterozygotes
  5. Why is population size important in hardy weinberg equilibrium as an evolutionary force?
    it is important because it reduces the change in alleles occurring by chance. In a small population, the frequency of alleles can change significantly with a small change in alleles. It increases the chances of genetic drift occurring in a small pop. IN a large population, genetic drift doesnt occur as much. Also there is much diversity in a large pop.
  6. the evaporation of water from the plant surface
  7. loss of liquids from the ends of vasular tissues at the margins of leaves
  8. What facilitates the transpirational pull?
    • cohesion
    • adhesion
  9. The upward transpirational pull on the xylem causes a __ (negative pressure) to form in the xylem, pulling the walls of the xylem inward. It also contributes to the lowering of water potential in the xylem.
  10. What is the purpose of the transpiration lab?
    • see how transpiration functions in the pull of water up the root to leaves
    • water potential in the plant
    • structure of plants
    • environmental conditions that increase or decrease transpiration
  11. Where did transpiration occur hte most? least?
    results based on mass?
    • fan
    • mist
    • the greatest decrease in mass showed that transpiration occurred the most
  12. How do you calculate percent change in mass of plants?
    • final-initial
    • initial x 100
  13. What was the independent and dependent variables? constants?
    • independent: time
    • dependent: amt of transpiration that took place (% change in mass)
    • constants: scale, type of plant, type of fan used
  14. Calculate the average rate of water loss per minute for each.
    per day.
    • first day- last day
    • 7200

    • first day- last day
    • 5
  15. Why did light cause an increase in transpiration?
    light: because the light introduced heat into the environment, it caused a pull of water from the roots to the leaves. ALso, because of teh light, potassium went into the guard cells, making them more negative, causing water to come into the guard cells causing a pull, which created a negative tension down to the roots.
  16. Why did dark cause a decrease in transpiration?
    without light, the guard cells are closed becuase of the lack of potassium in the guard cells. Therefore, no water is taken into the roots because nothing is being pulled out of the plant. Photosynthesis does not occur.
  17. Why did fan cause an increase in transpiration?
    The wind caused the environment around the plant to be drier. It decreased moisture around hte plant and facilitated evaporation causing a greater increase in transpiration.
  18. Why did the mist decrease transpiration?
    the mist decreased transpiration because the outside environment had a similar gradient; because they had a similar gradient, water diffused very slowly out of the plant due to the water that was present outside of the plant.
  19. How did each condition affect the gradient of water potential from the stem to leaf in the experimental plant?
    the light and fan affected the gradient of water potential because they decreased water potential. With less water otside the plant, the environment and air is dry. So water will transpire out of the plant since it moves to where it is less concentrated. The mist increased water potential because since there was an increase in humidity, the evaporation and transpiration decrased.
  20. What is the advantage and disadvantaeg to a plant of closed stamata wen water is in short supply?
    with water in short supply, closed stomata preserve the little amt of water left.Less transpiration occurs. However, gas exchange cannot occur. WIthout CO2, photosynthesis cannot occur
  21. What are some adaptations?
    • xerophytes
    • photosynthesis carried out in stems
    • CAM
    • small leaves
    • thick cuticles
    • store water in stems
  22. Why did you need to calculate the % water loss over the week instead of graphing the total amount of water lost each day?
    It corrects for any differences in mass. The plants did not have the same mass to begin with. Therefore, we wouldn't be sure if the mass changes. Also, there was more data present to allow for a more accurate experiment.
  23. How do you calculate hte change in mass per day?
    first- second day/ 1440

Card Set Information

Lab 8: Population Genetics and Evolution Lab 9: Transpiration
2011-04-23 17:23:02
AP Bio Eight Nine

AP Bio
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