Dihybrid Crosses

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[h] Mendel’s Dihybrid Crosses

[q] Mendel started, as always, with true-breeding varieties. But these varieties were true-breeding for two traits. In this case, a pea plant that was true-breeding for tall stems and purple flowers was bred with a plant that was true-breeding for short stems and white flowers.

 

Predict what the offspring will be. When you’ve written down your prediction, click “continue.”

[q] All of the offspring were tall, with purple flowers. Note, however, that the offspring are dihybrid (hybrid tall stem AND hybrid purple-flowered)

[q] Mendel’s next move was to carry out a dihybrid cross: He crossed his F₁ hybrids, producing the F₂s. Predict what happened.

Click after you’ve written down your prediction.

[q] So, here are the F₂ results that you get from an F₁ dihybrid cross.

• 9/16 have both dominant traits (in this case, tall stems AND purple flowers)
• 3/16 are dominant for the first trait and recessive for the second (tall stems, white flowers)
• 3/16 are recessive for the first trait and dominant for the second (short stems, purple flowers), and
• 1/16 are recessive for both traits (short stems, white flowers)

Explaining this 9:3:3:1 ratio got Mendel to his second great insight, his Principle of Independent Assortment. Let’s reconstruct Mendel’s thinking.

[q] In the previous card, you determined that each trait separately has a 3:1 ratio of dominant phenotype to recessive phenotype (3 tall to 1 short; and 3 purple-flowered to 1 white flowered). That’s the ratio we’d expect for a monohybrid cross. But how do we explain the 9:3:3:1 ratio that we see when we combine these phenotypes? It’s the result that we’d get if we did two independent Punnett squares and then combined the results, using what’s called the rule of multiplication. Start by completing the Punnett squares below.

Click to continue when you’re done.

[q] Here’s the solution:

Click when you’re ready.

[q] Now let’s talk about the rule of multiplication: when two events are independent, then the probability of them occurring together is the product of their independent probabilities. If I’m tossing a coin, the probability that four tosses will result in 4 heads is 1/2 x 1/2 x 1/2 x 1/2, or 1/16. Why? Because with each flip, the probability is 1/2, and these four events are independent.

Try this. Both the father and the mother in a family are heterozygous for cystic fibrosis (a recessive condition). They have two children. What’s the probability that both children will have cystic fibrosis? As you fill in the following blanks, you have to write out the numbers.

For the first child, the probability of being born with cystic fibrosis is one out of [hangman]. For the second child, it’s one out of [hangman]. That makes the probability for both children one over [hangman].

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[q] So that’s Mendel’s principle of independent assortment, derived from the laws of probability.

However, we can also explain what happens in a dihybrid cross by thinking about meiosis, where there’s also a process called independent assortment. Look at the diagram below, and write down a quick explanation of meiotic independent assortment. When you’re done, check it against the next slide.

[q] Here’s an explanation of meiotic independent assortment: when homologous pairs are lined up on the cell equator during metaphase 1, the way each pair lines up is independent of every other pair. For example, in the diagram below, the chromosome that originated from the mother is pink; the one from the father is blue (sorry for the stereotypes). In cell A, the maternal chromosome of one pair is facing up, as is the paternal chromosome of the second pair. In cell B, the maternal chromosome for the first pair is facing up, and the maternal chromosome for the second pair is facing up. After anaphase 1 and cytokinesis, this will result in different arrangements of chromosomes in the haploid gametes. Note that there are no other arrangements of cells A and B that produce different results from what’s shown in cells 1 through 4.

With two homologous pairs, you get four possible arrangements of chromosomes in the gametes, as shown below.

[q] Genes are located on chromosomes. So if chromosomes are independently assorting, then so are the genes that are on these chromosomes (as long as these genes are on different chromosomes: we’ll look at what happens when they’re not in the next tutorial). To make this as simple as possible, look at the diagram below.

In this germ cell (a diploid cell that’s going to make haploid gametes), there are two homologous pairs of chromosomes. Because this was a dihybrid cross, we’re going to put a dominant allele (T or P) on one member of each homolog, and a recessive allele (t or p) on the other homolog.

Each gamete has to have an allele for stem length, and an allele for flower color. So, what alleles can be in each of the four gametes?

Figure out the four different combinations, write them down, and then check your answer on the next card.

[q] Here’s the germ cell, and the four gametes that it could produce by independent assortment. Note you can use the FOIL algorithm from algebra (First, Outside, Inside, Last) to get these results.

In the next slide, you can use these gametes to create a Punnett square. Click when you’re ready.

[fx] No. Please try again.

[q] Here’s the table you produced on the previous slide.

	TP	Tp	tP	tp
TP	TTPP ♠	TTPp ♠	TtPP ♠	TtPp ♠
Tp	TTPp ♠	TTpp ♣	TtPp ♠	Ttpp ♣
tP	TtPP ♠	TtPp ♠	ttPP ♥	ttPp ♥
tp	TtPp ♠	Ttpp ♣	ttPp ♥	ttpp ♦

If you count up all the ♠ (tall, purple) you get [hangman] (please write out the number).

All the ♣ (tall, white) come to [hangman]

All the ♥ (short, purple) come to [hangman], and

All the ♦ (short, white) come to [hangman].

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