Midterm

Question 1: A model of meiotic drive

Meiotic drive is defined as the non-Mendelian production of gametes by heterozygotes.

That is, rather than producing equal numbers of "A" and "a" gametes, an Aa heterozygote produces a fraction, k, of "A" gametes and a fraction, 1-k, of "a" gametes.

Let p[t] equal the frequency of allele A among gametes of generation t. Assuming that these gametes unite at random and that there is no selection, the population of diploids at generation t will be at Hardy-Weinberg frequencies (p[t]^2 : 2p[t]q[t] : q[t]^2).

The following table gives the next generation of gametes produced by these diploid individuals in the presence of meiotic drive:

(a: 10 points) With meiotic drive, what will the frequency of the A gametes be in the next generation?

• p[t+1] = p[t]² + 2 p[t] q[t] k

(b: 15 points)When will the frequency of the A allele in the next generation be greater than the frequency in the current generation (i.e. when will p[t+1] be greater than p[t])?

• p[t+1] will be greater than p[t] when p[t+1]-p[t] > 0, i.e. when p[t]² + 2 p[t] q[t] k - p[t] > 0.

  Factoring out p[t], this gives that p[t] + 2 q[t] k - 1 > 0.

  But p[t]-1 = - q[t], so we get 2 q[t] k - q[t] > 0

  Factoring out q[t], this gives that 2 k - 1 > 0, i.e. k must be greater than 1/2 (there must be a greater than 50% chance that A is passed to the offspring of an Aa heterozygote).

Question 2: Graphical stability analysis

In the following graphs, p[t+1] has been plotted against p[t] for the diploid selection model in which mutation is also included (both from A to a and from a to A).

(a: 15 points) On these two graphs, place

• an X over locally unstable equilibria
• a small circle around equilibria that are locally but not globally stable
• a small square around equilibria that are globally stable.

Question 3: Comparing the spread of dominant versus recessive alleles.

The diploid selection model in continuous time can be solved directly in some cases of particular interest.

In the case of an allele (A) being favored that is dominant (WAA=1+s, WAa=1+s, Waa=1), the frequency of the A allele at time t is given by:

In the case of an allele (A) being favored that is recessive (WAA=1+s, WAa=1, Waa=1), the frequency of the A allele at time t is given by:

(a: 10 points) If the fitness of an AA individual is 10% higher than that of an aa individual (i.e. WAA=1.1 and s=0.1), how many generations does it take for the A allele to rise from a frequency of 0.01 to 0.5 when:

• A is dominant
  • When A is dominant, we use the first equation: Log[0.5/0.5] + 1/0.5 = Log[0.01/0.99] + 1/0.99 + 0.1 t.

    So t = (Log[0.5/0.5] + 1/0.5 - Log[0.01/0.99] - 1/0.99)/0.1 or 55.9 generations (the formula is in terms of natural log, we also accepted 29.9 generations if log base 10 was used).

• A is recessive
  • When A is recessive, we use the second equation: Log[0.5/0.5] - 1/0.5 = Log[0.01/0.99] - 1/0.01 + 0.1 t.

    So t = (Log[0.5/0.5] - 1/0.5 - Log[0.01/0.99] + 1/0.01)/0.1 or 1026 generations (the formula is in terms of natural log, we also accepted 1000 generations if log base 10 was used).

(b: 10 points) In 2-3 sentences, describe why selection is less effective in one of these cases, such that it takes a lot longer to accomplish the same change in allele frequency.

• Selection is less effective in the second case (A is recessive), because when the A allele is rare most individuals that carry the A allele are Aa heterozygotes not AA homozygotes. But when A is recessive, Aa heterozygotes have the same fitness as the common aa homozygote, so that there are very few fitness differences among the members of the population.

Question 4: Population growth model for Nerviosa complexa

Dr. McMuffin has been studying the species, Nerviosa complexa. She has found that while members of this species have plenty of food and are not resource limited, they get extremely agitated when they are crowded and their rate of reproduction falls off rapidly.

She has studied the number of offspring per parent as a function of the total population size and has observed that it decreases with the square of the population size unlike the normal logistic equation:

She decides, on the basis of this graph, that the population changes in size according to the equation:

Dr. McMuffin comes to you with help in analysing the dynamics of Nerviosa complexa.

(a: 7 points) What are the possible equilibrium population sizes?

• Either n = 0 or 1 - n²/K²=0.

  The second equation gives n² = K², so n = +/- Sqrt[K^{2] = +/- K.}

(b: 20 points) Analyse the stability of the only equilibrium point at which the population is not extinct. Specify when an equilibrium is locally stable or unstable and whether or not you expect to observe oscillatory behavior around the equilibrium.

• We only have to analyse the stability at n = +K.

  Expanding the recursion equation gives f(n) = n + r n - r n³/K².

  The derivative of this function with respect to the variable n is f'(n) = 1 + r - 3 r n²/K².

  This derivative evaluated at the equilibrium is f'(K) = 1 + r - 3 r K²/K² = 1 - 2 r.

  If r is negative, then this derivative is greater than one, so n=K is locally unstable.

  If 0 < r < 1/2, then the derivative falls between 1 and 0, implying a non-oscillatory approach to equilibrium.

  If 1/2 < r < 1, then the derivative falls between 0 and -1, implying an oscillatory approach to equilibrium.

  If 1 < r , then the derivative falls below -1, implying that the equilibrium is unstable and fluctuations around the equilibrium are expected.

(c: 5 points) Based on this analysis, would you predict that the dynamics for Nerviosa complexa would become chaotic for lower values of r, the same values of r, or higher values of r than in the standard logistic model? (Circle one)

• In the regular logistic model, oscillations aren't seen until r>2, implying that the dynamics for Nerviosa complexa are undergoing period doubling sooner (when r>1) and suggesting that chaos will be seen for lower values of r.

  An alternative way of answering would be to say that Nerviosa complexa shows much stronger dependence on the density of the population (the number of offspring per parent goes down as n²). Because chaos arises from overshooting and then undershooting the equilibrium, you might expect the dynamical system of Nerviosa complexa to enter chaos for lower values of r.

(d: 8 points) Dr. McMuffin also says that the intrinsic rate of growth of Nerviosa complexa is quite low. You then tell her that the discrete equation may be approximated by a differential equation that can be solved generally. Write down this differential equation.

• We can use the definition of a derivative to give us that dn/dt = Limit[(n[t+h] - n[t])/h] as h -> 0. For low values of r, this can be approximated by setting h=1, in which case we have dn/dt = (n[t+1] - n[t])/1 = r n (1-n²/K²). However, this only works if r is small enough so that the change over small time periods can be predicted by change over one generation.

EXTRA CREDIT QUESTION

(a: 10 points) Solve the differential equation in part (d) of question 4 by the method of separation of variables. [Hint: Mathematica tells you that Integrate[1/(n*(1-n²/k²)),n] equals Log[n] - Log[-k² + n²]/2.]

(WARNING: This part is tricky. Attempt only if you have enough time.)

• dn/dt = r n (1-n²/K²)

  Separating the variables and integrating both sides: Integrate[dn/(n (1-n²/K²))] = Integrate[r dt]

  Integrating: Log[n] - Log[-k² + n²]/2 = r t + c

  Multiply everything by 2: 2 Log[n] - Log[-k² + n²] = 2 r t + c

  Rearrange the first log term: Log[n²] - Log[-k² + n²] = 2 r t + c

  Gather the log terms: Log[n²/(-k² + n²)] = 2 r t + c

  Exponentiate both sides: Exp[Log[n²/(-k² + n²)]] = Exp[2 r t + c]

  Simplify: n²/(-k² + n²) = Exp[2 r t] C

  At t = 0, n0²/(-k² + n0²) = Exp[2 r 0] C = C, so

  n²/(-k² + n²) = Exp[2 r t] n0²/(-k² + n0²)

  Multiply both sides by the denominator on the right: n² = (-k² + n²) Exp[2 r t] n0²/(-k² + n0²)

  Gather n² terms: n² (1 - Exp[2 r t] n0²/(-k² + n0²)) = -k² Exp[2 r t] n0²/(-k² + n0²)

  Divide both sides by the factor on the right: n² = -k² Exp[2 r t] n0²/(-k² + n0²)/(1 - Exp[2 r t] n0²/(-k² + n0²))

  Simplify a bit: n² = k² Exp[2 r t] n0²/(k² - n0²)/(1 + Exp[2 r t] n0²/(k² - n0²))

  Square-root both sides: n = Sqrt[k² Exp[2 r t] n0²/(k² - n0²)/(1 + Exp[2 r t] n0²/(k² - n0²))]

  Tah-dah!

  Does it check out? When t gets large, it goes to K. When t=0, it goes to n0!