Question

Mendelian GeneticsWhen Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 929 peas, with 705 of them having red flowers. If we assume, as Mendel did, that under these circumstances, there is a 3/4 probability that a pea will have a red flower, we would expect that 696.75 (or about 697) of the peas would have red flowers, so the result of 705 peas with red flowers is more than expected.

a.If Mendel’s assumed probability is correct, find the probability of getting 705 or more peas with red flowers.

b.Is 705 peas with red flowers significantly high?

c.What do these results suggest about Mendel’s assumption that 3/4 of peas will have red flowers?

Short answer

a. Probability of getting 705 or more peas with red flowers is 0.2776.

b. 705 peas with red flowers is not significantly high.

c. Since it is not highly significant, this experiment does contradict Mendel’s theory.

Step-by-step solution

Given information

The number of peas in one sample of offspring is recorded.

The given sample size\(n = 929\)and probability of success \(p = 0.75\left( {\frac{3}{4}} \right)\).

Then,

\(\begin{aligned}{c}q = 1 - p\\ = 1 - 0.75\\ = 0.25\end{aligned}\)

Check the requirement

From the given information,

\(\begin{aligned}{c}np = 929 \times 0.75\\ = 676.75\\ > 5\end{aligned}\)

And

\(\begin{aligned}{c}nq = 929 \times 0.25\\ = 232.25\\ > 5\end{aligned}\)

Here both\({\bf{np}}\)and\({\bf{nq}}\)are greater than 5. Hence probabilities from a binomial

probability distribution can be approximated reasonably well by using a normal distribution.

Mean and standard deviation for normal distribution

The mean value is,

\(\begin{aligned}{c}\mu = np\\ = 929 \times 0.75\\ = 696.75\end{aligned}\)

The standard deviation is,

\(\begin{aligned}{c}\sigma = \sqrt {npq} \\ = \sqrt {929 \times 0.75 \times 0.25} \\ = 13.20\end{aligned}\)

Compute the probability

a.

The probability of getting 705 or more peas with red flowers is expressed as using\(P\left( {X \ge n} \right)\;{\rm{use}}\;P\left( {X > n - 0.5} \right)\)

Thus

\(\begin{aligned}{c}P\left( {X \ge 705} \right) = P\left( {X > 705 - 0.5} \right)\\ = P\left( {X > 704.5} \right)\end{aligned}\)

Thus, the probability is expressed as \[P\left( {X > 704.5} \right)\].

The z-score for\(x = 704.5\),\(\mu = 696.75,\sigma = 13.20\) is as follows:

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{704.5 - 696.75}}{{13.20}}\\ = 0.59\end{aligned}\)

The z-score is 0.59.

Using the standard normal table, the cumulative probabilities are obtained for 0.59 as, 0.7224.

Substituting the value,

\(\begin{aligned}{c}P\left( {Z > 0.59} \right) = 1 - 0.7224\\ = 0.2776\end{aligned}\)

Thus, the probability is 0.2776.

Define significantly high

b.

The event of getting 705 peas with red flowers would be significantly high if the probability of getting value 705 or above is lesser than 0.05.

In this case, 0.2776 > 0.05.

Therefore, it is not significantly high.

Discuss the significance of the results

c.

As the value is not significantly high, it can be inferred that the chances of getting 705 or more peas do not suggest strong evidence against the claim that 0.75 proportion of peas will have red flowers.