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The relationship between the frequency of an allele A, $x$, at a genetic locus in a diploid population and the fitness of a population with this frequency of allele A, $w$, is described by the function $w=ax^2+x(b-x)+c(b-x)^2$ . The aims are (a) ti simplify the algebraic expression, (b) calculate the fitness of a population with a given allele A frequency, and (c) calculate the allele A frequency when the fitness of the population is given.

", "licence": "None specified"}, "statement": "

Let $x$ represent the frequency of an allele A at a genetic locus in a diploid population. The fitness, $w$, of a population with this frequency of allele A is given by:

\\[ \\simplify{w={a}x^2+x({b}-x)+{c}({b}-x)^2} \\]

", "advice": "

a) We need to open the brackets and collect like terms:

\\[ \\begin{split} w &= \\simplify{{a}x^2+x({b}-x)+{c}({b}-x)^2} \\\\ \\implies w &= \\simplify[!basic]{{a}x^2+x({b}-x)+{c}({b}-x)({b}-x)}\\\\ \\implies w &= \\simplify [unitFactor, !basic] {{a}x^2+x({b}-x)+{c}({b}^2-{b}x-{b}x+x^2)} \\\\ \\implies w &= \\simplify {{a}x^2+x({b}-x)+{c}({b}^2-{b}x-{b}x+x^2)} \\\\ \\implies w &= \\simplify[unitFactor, !collectNumbers] {{a}x^2+{b}*x-x^2+{c}*{b^2}-{c}*{2*b}x+{c}x^2} \\\\ \\implies w &=\\simplify[all, !noLeadingMinus] {{a}x^2+{b}*x-x^2+{c}*{b}^2-2*{c}*{b}x+{c}x^2} \\end{split}\\]

b) We substitute $x=\\var{x_1}$ in the simplified formula:

\\[w=\\simplify[unitfactor, !basic]{{coef_a}*{x_1}^2+{coef_b}*{x_1}+{coef_c}} \\\\ \\implies w=\\simplify [unitfactor, !basic]{{coef_a}*{x_1^2}+{coef_b}*{x_1}+{coef_c}} \\\\ \\implies w=\\var{w_ans_rounded} \\text{ (2 d.p.)}\\]

Note: we get the same result regardless of which version of the formula we use. Using the simplified formula however is quicker.

c) We need to substitute $w=\\var{w_1}$ and solve the quadratic equation:

\\[ \\simplify[!noLeadingMinus]{{w_1}={coef_a}*x^2+{coef_b}*x+{coef_c}} \\\\ \\implies \\simplify[!noLeadingMinus]{{w_1}-{coef_a}*x^2-{coef_b}*x-{coef_c}=0} \\\\ \\implies \\simplify{{w_1}-{coef_a}*x^2-{coef_b}*x-{coef_c}=0}\\]

Using the quadratic formula

\\[ x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\]

We get the solutions:

\\[ \\begin{split} x_1 &\\,= \\frac{\\simplify{-{co_b}+sqrt({co_b}^2-4*{co_a}*{co_c})}}{\\simplify{{2*co_a}}} \\qquad \\text{and} \\qquad x_2 &\\,= \\frac{\\simplify{-{co_b}-sqrt({co_b}^2-4*{co_a}*{co_c})}}{\\simplify{{2*co_a}}} \\\\\\\\ &\\,= \\var{ansx_1_rounded},\\quad &\\,=\\var{ansx_2_rounded}. \\end{split}\\]

Taking into consideration the context of the problem, we need to reject one of the two answers if it is negative. Thus, the final answer is $x=\\var{ans_c}$.

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Write $w$ in the form $ax^2+bx+c$.

$w=$[[0]]

Calculate the fitness of a population with allele frequency $x=\\var{x_1}$.

$w=$[[0]]

Give your answer rounded to 2 decimal places.

Find the allele frequency, $x$, of a population with fitness $\\var{w_1}$.

$x=$[[0]]

Give your answer rounded to 2 decimal places, if needed.

The question includes a quadratic graph depicting the relationship between the frequency of an allele A at a genetic locus in a diploid population and the fitness of a population with this frequency of allele A. The aim is to estimate the maximum and minimum fitness of the population and the corresponding frequency of allele A.

", "licence": "None specified"}, "statement": "

The following graph depicts the relationship between the frequency, $x$, of an allele A at a genetic locus in a diploid population and the fitness, $w$, of a population with this frequency of allele A.

{geogebra_applet('https://www.geogebra.org/m/rtngjns4', defs)}

", "advice": "

{geogebra_applet('https://www.geogebra.org/m/xw4jjqxa',defsad1)}

Note: you can zoom in and out of the figure as needed.

a) We first need to locate the highest point on the graph. From that point, we can draw a line perpendicular to the vertical axis to estimate the maximum fitness and a line perpendicular to the horizontal axis to estimate the corresponding frequency of allele A.

Here, a good estimate could be:

maximum fitness, $w_{max}=\\var{max}$
frequency of allele A, $x=\\var{xmax}$.

b)We first need to locate the lowest point on the graph. From that point, we can draw a line perpendicular to the vertical axis to estimate the minimum fitness and a line perpendicular to the horizontal axis to estimate the corresponding frequency of allele A.

Here, a good estimate could be:

minimum fitness, $w_{min}=\\var{min}$
frequency of allele A, $x=\\var{xmin}$.

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Use the graph to find the maximum possible fitness and the value of $x$ at which this maximum is attained.

Maximum fitness, $w_{max}=$ [[0]].
Frequency of allele A, $x=$ [[1]].

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Use the graph to find the minimum possible fitness and the value of $x$ at which this maximum is attained.

Minimum fitness, $w_{min}=$ [[0]].
Frequency of allele A, $x=$ [[1]].

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The proportion of the sodium carbonate, $p$, which has dissolved by time $t$ seconds is given by the formula $ p=\\frac{bt-at^2}{c}$. The aim is to calculate the proportion of sodium carbonate in a solution at a given time and vice versa.

", "licence": "None specified"}, "statement": "

A small volume of sodium carbonate is added to a beaker of water. The proportion of the sodium carbonate, $p$, which has dissolved by time $t$ seconds is given by:

\\[p=\\frac{\\var{b}t-\\var{a}t^2}{\\var{c}}\\]

", "advice": "

a) We need to substitute $t=\\var{t_1}$
\\[ p=\\frac{\\var{b}\\times\\var{t_1}-\\var{a}\\times \\var{t_1}^2}{\\var{c}}\\\\ \\implies p=\\frac{\\var{b}\\times\\var{t_1}-\\var{a}\\times \\var{t_1^2}}{\\var{c}} \\\\ \\implies p=\\frac{\\simplify{{b}*{t_1}}-\\simplify{{a}*{t_1^2}}}{\\var{c}} \\\\ \\implies p=\\frac{\\simplify{{b}*{t_1}-{a}*{t_1^2}}}{\\var{c}} \\\\ \\implies p=\\simplify{{b*t_1-a*t_1^2}/{c}} ~~~~ \\text{or} ~~~~ \\var{ans_a_rounded}~\\text{(2d.p.)} \\]

b) We first need to transform the percentage value which we were given into a proportion written as a decimal.

\\[ \\var{p_1*100}\\%=\\var{p_1}\\]

Then, substituting $p=\\var{p_1}$ we get the following equation to solve:

\\[ \\var{p_1}=\\frac{\\var{b}t-\\var{a}t^2}{\\var{c}} \\\\ \\implies \\var{c}\\times
\\var{p_1}=\\var{b}t-\\var{a}t^2 \\\\ \\implies \\simplify[unitFactor]{{a}t^2-{b}t+{c*p_1}}=0\\]

Using the quadratic formula

\\[ t=\\frac{-b\\pm \\sqrt{b^2-4a c}}{2a}\\]

we get the solutions:

\\[ t_1=\\var{anst_1_rounded} \\\\ t_2=\\var{anst_2_rounded}\\]

However, we are solving the equation to find a time which is $0\\leq p\\leq \\var{tmax}$. Therefore, the final answer is only $t=\\var{ans_t}$.

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Calculate the proportion of the sodium carbonate which has dissolved after $\\var{t_1}$ seconds.

$p=$ [[0]]

Give your answer rounded to 2 decimal places.

Calculate the time taken for $\\simplify{{p_1}*100}$ % of sodium carbonate to dissolve.

$t=$ [[0]] sec.

Give your answer rounded to 2 decimal places.

Estimating the proportion of sodium carbonate in a solutionat a specific timepoint and vice versa, depicted as a quadratic graph.

", "licence": "None specified"}, "statement": "

A small volume of sodium carbonate is added to a beaker of water. The proportion of the sodium carbonate, $p$, which has dissolved by time $t$ seconds is illustrated in the graph below for $0\\leq t\\leq \\var{tmax}$.

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a) We first need to find the point on the time axis for $\\var{t_1}$ seconds. From that point, we draw a line vertically to find a point on the graph. From the point on the graph, we need to draw a horizontal line, and then estimate the value of $p$.

{geogebra_applet('https://www.geogebra.org/m/wnracghs', defsad)}

Here, a good estimate could be $p=\\var{ans_a_rounded} $.

b) We first need to estimate the point where the $\\var{p_1*100}$% would lay on would lie on the p-axis. That is $ \\frac{\\var{p_1*100}}{100}=\\var{p_1}$. From here, we draw a line horizontally to find a point on the graph. From the point on the graph, we draw a vertical line, and then estimate the value of $t$.

{geogebra_applet('https://www.geogebra.org/m/wnracghs', defsad1)}

Here, a good estimate could be $\\var{ans_t}$ seconds.

c) At 7 seconds, the proportion of sodium carbonate which has dessolved is $p=1$ or, if transformed to a percentage, $100$%. That means all the sodium carbonate will have dissolved completely at $t=7$ seconds and the quadratic graph cannot describe the situation beyond the 7 seconds.

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Use the graph to estimate the proportion of sodium carbonate which has dissolved after $\\var{t_1}$ seconds.

$p=$ [[0]]

Give your answer as a decimal, if needed, rounded to 2 decimal places.

Use the graph to estimate the time taken for $\\simplify{{p_1}*100}$ % of the sodium carbonate to dissolve.

$t=$ [[0]] sec.

Give your answer rounded to 2 decimal places.

Why is this model not appropriate when $t\\geq \\var{tmax}$. You can choose only one of the answers.

[[0]]

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