Question

The seeds of the garden pea (Pisum sativum) are either yellow or green. A certain cross between pea plants produces progeny in the ratio 3 yellow : 1 green. ${}^{16}$ If four randomly chosen progeny of such a cross are examined, what is the probability that (a) three are yellow and one is green? (b) all four are yellow? (c) all four are the same color?

Short answer

The probability is (a) 27/64 for three yellow and one green, (b) 81/256 for four yellow, and (c) 41/128 for all four progeny being the same color.

Step-by-step solution

Understanding the Genetic Cross

The given ratio for the progeny of pea plants is 3 yellow to 1 green, which is a classical Mendelian inheritance pattern for a single gene with dominance. This ratio suggests that we are observing a monohybrid cross where yellow seeds are dominant over green. To find probabilities involving specific outcomes, we use the binomial probability formula.

Calculate Probability for Part (a)

For part (a), we want the probability of having three yellow and one green among four offspring. Using the binomial probability formula, which is P(X=k) = (n choose k)p^k(1-p)^(n-k), where n is the number of events, k is the specific number of successes one is solving for, p is the probability of success, and (n choose k) is the binomial coefficient, we can calculate this. Here, n=4 (four progeny), k=3 (three are yellow), p=3/4 (probability of yellow), and q=1/4 (probability of green). The calculation is P(3 yellow, 1 green) = (4 choose 3)(3/4)^3(1/4)^1.

Calculate Probability for Part (b)

For part (b), the probability that all four are yellow is found using the same formula with k=4 since all four progeny need to be yellow. Thus, P(X=4) = (4 choose 4)(3/4)^4(1/4)^0. Since any number to the zero power is 1, this simplifies the calculation significantly.

Calculate Probability for Part (c)

For part (c), we are looking for the probability that all four offspring are the same color. This is the sum of the probabilities of all four being yellow or all four being green. We already calculated the probability of all four yellow in Step 3. For all green, we use n=4, k=4, p=1/4 since green is now the 'success' we're counting. Then we add the two probabilities together: P(4 yellow) + P(4 green).

Solve and Summarize

Now, perform the actual calculations: For part (a), (4 choose 3)(3/4)^3(1/4)^1 = 4 * (27/64) * (1/4) = 27/64. For part (b), P(4 yellow) = (4 choose 4)(3/4)^4(1/4)^0 = 1 * (81/256) = 81/256. For part (c), we add the probability of all yellow and all green. P(4 green) is (4 choose 4)(1/4)^4(3/4)^0 = 1/256. So, P(all same color) = P(4 yellow) + P(4 green) = 81/256 + 1/256 = 82/256 = 41/128 after simplifying the fraction.

Key concepts

Monohybrid Cross

Imagine walking through a garden of pea plants and noticing a mix of yellow and green seeds. This color difference arises from a fascinating genetic principle: a monohybrid cross. In such a cross, we're dealing with just one gene that controls the color of the seeds. The yellow color is determined by a dominant allele, represented by a capital letter (Y), and the green by a recessive allele, shown as a lowercase letter (y).

When two pea plants are crossed, the offspring, or the 'progeny', have different combinations of these alleles. Mendel's law of segregation tells us that allele pairs separate during gamete formation and randomly unite at fertilization. So, if we start with parents that are heterozygous (Yy) for the seed color gene, they can produce offspring that are yellow (YY or Yy) or green (yy) with the famous 3:1 ratio in a monohybrid cross.

Binomial Probability Formula

To predict the likelihood of different combinations of seed colors among the offspring, we turn to the binomial probability formula. The 'binomial' part means we're dealing with two possible outcomes: yellow or green. Here's the formula at work:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

That might look intimidating, but let's break it down. $P(X=k)$ gives us the probability of getting exactly 'k' successes (in this case, yellow seeds) in 'n' trials (the number of offspring we're examining). $(\genfrac{}{}{0ex}{}{n}{k})$ is the binomial coefficient, which calculates how many ways 'k' successes can occur in 'n' trials, and 'p' is the probability of one success. This formula is like a recipe, telling us how to combine these ingredients to predict patterns of inheritance in genetics.

Genetic Dominance

In the colorful world of pea plants, not all alleles are created equal. Some shout louder than others. This is the essence of genetic dominance. An allele for yellow seeds, the one shouting louder, masks the expression of the allele for green seeds, which is a wallflower in the genetic conversation. This dominant allele (Y) requires only one copy to be present for its trait (yellow seeds) to be expressed in the plants. On the other hand, a green seed color caused by the recessive allele (y) only appears when there are two copies, since there's no dominant allele to overshadow it. This relationship between alleles determines the visible traits, or phenotypes, and follows Mendel's laws of inheritance.

Probability Calculations

Now, how do we apply this to gardening or genetics homework? Through probability calculations, which are a way of predicting the occurrence of certain traits. Remember our pea plants? Let's say we want to know the chances of plucking four yellow seeds from our next generation of plants. Using our binomial formula and knowledge of genetic dominance, we can calculate this probability. We'd need to know the total number of seeds we're looking at (n) and the number of yellow seeds we're interested in finding (k), along with the probability of finding a yellow seed (p) on any single pick. Combining these numbers in our formula gives us a clear mathematical route to understanding the likelihood of our garden producing a certain pattern of seed colors.