5 Steps to a 5 AP Biology, 2014-2015 Edition

308 ❯ STEP 5. Build Your Test-Taking Confidence

6. In Earth’s early history, the evolution of photo-
   synthesis in simple cells occurred before the evo-
   lution of more complex cells. (maximum 4 points
   for entire question)
   
   
   • Mentioning that photosynthesis releases oxygen
     as a by-product. (1 point)
   
   
   • Mentioning that photosynthesis led to an
     increase in atmospheric oxygen. (1 point)
   
   
   • Mentioning that the presence of more oxygen
     in the atmosphere allowed the evolution of
     cellular respiration. (1 point)
   
   
   • Mentioning that oxygen allowed cells to gen-
     erate more energy and grow larger and more
     complex. (1 point)
   
   
   • Mentioning that the first photosynthetic cells
     were prokaryotic. (1 point)
   
   
   • Mentioning that eukaryotic cells could not
     evolve until there was a higher level of atmos-
     pheric oxygen. (1 point)
   
   
   
   


7. What evidence supports the theory that chloro-
   plasts and mitochondria are evolved from
   prokaryotic cells? (maximum 4 points for entire
   question)
   
   
   • Mentioning that chloroplasts and mitochon-
     dria have their own DNA. (1 point)
   
   
   • Mentioning that chloroplast and mitochon-
     drial DNA consist of a single, circular molecule
     (like bacterial DNA). (1 point)
     
     
     • Mentioning that chloroplast and mitochondrial
       DNA are not associated with histones (like
       bacterial DNA). (1 point)
     
     
     • Mentioning that chloroplasts and mitochon-
       dria replicate by a process similar to prokary-
       otes. (1 point)
     
     
     • Mentioning that the inner membranes of both
       organelles have enzymes homologous to those
       found in prokaryotes. (1 point)
     
     
     
     
   
   
   
   


8. You are asked to estimate if a certain species of
   plant could live in a salt marsh. (maximum 4
   points for entire question)

• Calculate Ψplant cell:
  Ψ=pressure potential +solute potential solute
  potential of plant cell =Ψs=–iCRT=
  –(1)(0.08 M)(.00831)(273 + 12)=
  –0.189MPa (1 point)
  Ψ=–1.2 MPa (pressure potential) +–0.189
  MPa (solute potential) = –1.39 MPa
  (1 point)


• Mentioning that the plant cell’s water poten-
  tial (–1.39 MPa) is higher than that of the soil
  (–2.2 MPa). (1 point)


• Mentioning that water would flow out of the
  plant cell (hypotonic) into soil (hypertonic).
  (1 point)


• Mentioning that the plant cell would not
  survive. (1 point)