"activity" - answer key

L112 – Learning Group Week #1
Interpreting Graphs
1. How much does camouflage affect predation on mice by owls with and without
How the experiment was done:
Pairs of mice with different coat colors, one light brown and one dark brown, were released
simultaneously into an enclosure that contained a hungry owl. The researcher recorded the color
of the mouse that was first caught by the owl. If the owl did not catch either mouse within 15
minutes, the test was recorded as a zero. The release trials were repeated multiple times in
enclosures with either a dark-colored soil surface or a light-colored soil surface. The presence or
absence of moonlight during each was recorded.
Graph A shows data from the light-colored soil enclosure and graph B from the dark-colored
enclosure, but all of the other respects of the graphs are the same.
Using the bar graph above, answer the following questions:
a. What are the independent variables (what is being tested) in these graphs?
The independent variables for each graph are the coat color of the mice (light or dark
brown) and the presence or absence of moonlight (full moon or no moon). Taking both
graphs together, a third independent variable is the color of soil in the enclosure. (2pt)
b. What is the dependent variable (what is being measured – the response to the tests)?
The dependent variable is the amount of predation, measured as the number of mice
caught. (1pt)
c. Is a dark brown mouse on dark-colored soil more likely to escape predation under a full
moon or with no moon? A light brown mouse on light-colored soil? Explain. (1pt)
Under a full moon (12 were caught vs. 20 under no moon).
Under no moon (11 were caught vs. 18 under a full moon).
d. Under which condition would a dark brown mouse be most likely to escape predation at
night? A light brown mouse? (1pt each)
Dark soil field with a full moon. // Light soil with no moon
e. Which conditions are most deadly for both light brown and dark brown mice? (1pt)
Being on the contrasting soil is most deadly for both colors of mice.
2. Does delayed breeding increase the number of years a bird can breed?
Siberian Jays are birds that sometimes stick around and help their parents raise future offspring
(their siblings or half-siblings) instead of going off on their own to breed. Researchers were
interested in whether or not birds may gain an advantage in terms of their own lifetime
reproductive success (# of offspring produced), if they delayed the age at which they left their
parents and started to breed.
gray bars: delayed
white bars: no delay
3pts – one pt each –
Using the bar graph above, answer the following questions:
a. In this study, the researchers did not do any manipulations of variables instead they made
observations of a natural population of Siberian Jays. What are the independent and
dependent variables?
independent: delayed breeding or no delayed breeding
dependent: # of years of breeding
b. What does the graph above tell us about the advantage (or the lack) of delaying dispersal
and reproduction?
birds that delayed their own breeding had more years of breeding & hence potentially
more offspring over their lifetime.
c. How would you quantify the data above for a statistical comparison?
Averages – the average # of years individuals were able to breed if they delayed breeding
vs if they did not delay
NOTE: would be good to measure the # of offspring they had in each year of breeding
too. (which they did, but not shown on this graph)
Review of Chemistry:
1. Use the following list of atoms and their associated atomic number to answer the questions
Hydrogen – 1
Carbon – 6
Nitrogen – 7
Oxygen – 8
(a) How many possible binding sites does a Nitrogen atom have? 3 (1pt)
(b) How many hydrogen atoms can bind to one Carbon atom? 4 (1pt)
(c) Assuming a hydrocarbon chain (carbons strung together with hydrogens sticking off),
diagram how three carbons and 6 hydrogens would be bound together. Use a solid line to
represent a covalent bond
The absolute location of the double bond is not critical (though it has to be between two
carbons. The end carbon within the double bond will have two hydrogens attached while
the middle carbon will have only one hydrogen attached. The carbon atom not within the
double bond will have three hydrogens attached. (1pt)
(d) Diagram out the following molecules using the “mini solar system” model to represent the
electrons within the electron shells circulating around the nucleus. (2pt – each)
H2 O
NOTE: it is better to have all four electrons in a row in the double bond but it was hard for
me to draw it like that and keep it in the lines with the powerpoint I was using. J
2. Complete the following chemical formulas so that the equation is in equilibrium.
C6H12O6 + 6O2 ⇔ 6CO2 + 6H2O
CO2 + H2O ⇔ H2CO3 ⇔ H+ + HCO33. Given the following equation which type of bond would you predict is holding the product of
the reaction together? Explain your reasoning (1pt)
Na+ + Cl- -----> NaCl
Ionic Bond – two ions (one cation and one anion) come together to form a compound that is
electrically neutral.