AP Biology Practice Exam — Free

Work through what GTP binding and GTP hydrolysis each do in a G protein cycle. In the resting state the alpha subunit holds GDP and is inactive. When epinephrine binds the receptor, the receptor causes the alpha subunit to release GDP and bind GTP; the GTP-bound alpha subunit is the active form that stimulates adenylyl cyclase. The G protein is also its own timer: its built-in GTPase activity hydrolyzes the bound GTP to GDP after a short time, which switches the signal off. Washing away the ligand only stops new activations; G proteins that are already active are shut down by hydrolysis.

In the mutant, GTP binding (activation) is normal, so the pathway turns on as usual. What is lost is the off switch. The alpha subunit stays locked in the GTP-bound, active state, so adenylyl cyclase keeps converting ATP to cAMP, protein kinase A stays active, and the downstream enzymes keep breaking down glycogen even though the hormone is no longer present.

The signal cannot be terminated: cAMP stays elevated and glycogen breakdown continues after epinephrine is removed (option A).

Option B is the most common confusion; it mixes up the two steps. Hydrolysis is what turns the G protein off, not on, so blocking it cannot prevent activation. Option C is wrong because the mutation is in the G protein, not the receptor, so ligand binding is unchanged. Option D ignores that hydrolysis is the only way the active alpha subunit returns to its GDP-bound state; without it the response is prolonged, which is the same mechanism by which cholera toxin causes persistent cAMP signalling in intestinal cells.

The logistic model scales the exponential growth rate by the fraction of the carrying capacity that is still unused:

Identify each quantity: $N = 400$ is the current population, $K = 1000$ is the carrying capacity, and $r_{max} = 0.50$ per year is the maximum per-capita growth rate. The factor $(K - N)/K$ is the proportion of the environment's capacity still available; it equals 1 when the population is tiny and 0 when the population reaches $K$.

Substitute:

Interpretation: with no limit the population would grow by $0.50\times400 = 200$ rabbits per year, but because it is already at 40 % of carrying capacity, competition for resources reduces the realized growth to 60 % of that figure. The growth rate of a logistic population is highest at $N = K/2 = 500$; above that it falls, reaching zero at $N = K$.

Common mistakes are to drop the parentheses (computing $0.50\times400\times1000 - 400$) or to use $N/K$ instead of $(K - N)/K$, which gives 80 rabbits per year.

Growth rate: 120 rabbits per year.

(a) The null hypothesis is that the observed ratio does not differ from the expected one: the two genes are unlinked and assort independently, so the dihybrid cross PpTt × PpTt produces offspring in a 9 : 3 : 3 : 1 phenotypic ratio. Expected counts are the ratio fractions times the total of 160: purple tall $\tfrac{9}{16}(160) = 90$, purple short $\tfrac{3}{16}(160) = 30$, white tall $30$, and white short $\tfrac{1}{16}(160) = 10$. These sum to 160, as they must.

(b) Chi-square sums, over every class, the squared difference between observed ($o$) and expected ($e$) divided by the expected:

(c) Degrees of freedom equal the number of phenotype classes minus one: $4-1 = 3$. The calculated value, 1.31, is far below the critical value of 7.815, so the probability of seeing deviations at least this large by chance alone is greater than 0.05. The null hypothesis is not rejected: the data are consistent with independent assortment, so there is no evidence that the flower-colour and stem-height genes are linked. Note that failing to reject does not prove the genes are unlinked; it means the data give no reason to doubt it. A chi-square above 7.815 would have suggested linkage or some other departure from the 9 : 3 : 3 : 1 model.

(a) Expected 90 : 30 : 30 : 10; (b) $\chi^2 = 1.31$; (c) not rejected; the data are consistent with independent assortment.