1. Introduction

In the previous genetics tutorials, we’ve looked at variations on crosses that involved a single gene, with two alleles (or three, in the case of blood type). But what happens when two (or more) genes are being transmitted at the same time?

2. From Segregation of Alleles (Mendel’s 1st principle) to Dihybrid Crosses

The table below shows seven of the traits in garden peas that Gregor Mendel, the founder of genetics, experimented with as he was figuring out the principles of genetics in the 1850s and 60s.

Mendel’s principle of segregation of alleles came out of his experiments with monohybrid crosses. In these crosses, Mendel crossed true breeding varieties that differed in one trait to create a hybrid. For example, if Mendel wanted to create a hybrid for seed color, he’d cross a true breeding plant with round seeds with a true breeding plant that produced wrinkled seeds. The result would be a hybrid plant with round seeds (because round is the dominant allele).

We’ve seen that a monohybrid cross produces offspring in a 3 to 1 ratio, with 3 offspring showing the dominant trait for every one showing the recessive trait. Here’s the Punnett square. We’ll use the letter “R” to represent the allele for round seeds (the dominant trait) and “r” to represent the allele for wrinkled seeds, which is recessive.

R r
R RR Rr
r Rr rr

75% of the offspring have genotypes RR or Rr, giving them round seeds. 25% have wrinkled seeds (genotype rr).

Why is this called the principle of segregation? The name reflects Mendel’s insight that parents possess two alleles for a trait. When that parent creates offspring, he/she/it only passes on one of its two alleles. The alleles separate (or segregate) during gamete formation.

Along with figuring out the principle of segregation, Mendel also experimented with dihybrid crosses: crosses in which individuals were hybrid for two traits, such as stem height and flower color.

Follow what Mendel did through the interactive reading below.

[qwiz qrecord_id=”sciencemusicvideosMeister1961-dihybrid crosses, Interactive reading”]

[h]Mendel’s Dihybrid Crosses: Interactive Reading.

[q]We’ll start with seeing how Mendel produced his dihybrids. He 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.

Both parents are true breeding.

 

Predict what the offspring will be. When you’ve written your prediction in your student learning guide, 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)

“P” stands for parental generation. The F1 generation is the P generation’s offspring. All the F1s are hybrid.

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

Click after you’ve written down your prediction.

[q labels = “right”]Here’s what happened. There were four phenotypes. And, once Mendel analyzed his data, he found that they occured in predictable ratios. Try to predict which ratio goes with which phenotype. Don’t worry if you don’t know the answer. Just guess and drag these fractions until they’re in the right place.

 

[l]1/16

[fx] No. Please try again.

[f*] Correct!

[l]3/16

[fx] No. Please try again.

[f*] Excellent!

[l]9/16

[fx] No. Please try again.

[f*] Correct!

[q]So, here’s what you get from a 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 labels = “right”]Dihybrid cross results:

  • Tall stem, purple flowers: 9/16
  • Tall stem, white flowers: 3/16
  • Short stem, purple flowers: 3/16
  • Short stem, white flowers: 1/16

Let’s look at each characteristic separately.

If there were 16 offspring, you could expect ______ to have tall stems; and  ______ to have short stems

Reducing that that to a ratio you’d expect  ______  with tall stems for every  ______  with short stems.

 

Now, do the same for flower color. Out of 16, you’d expect ______ with purple flowers; and ______ with white flowers.

That’s a ratio of ______  with purple flowers to ______  white flowers.

[l]3/4

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]1/4

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]12

[fx] No. Please try again.

[f*] Great!

[l]4

[fx] No. Please try again.

[f*] Good!

[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 in your student learning guide.

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].

[c]four

[c]four

[c]sixteen

[q labels = “top”]Likewise, if we conclude (as Mendel did) that the trait for stem length operates independently from the trait for flower color, then the probability of a plant being tall AND purple flowered is the product of a plant being tall (3/4), multiplied by the probability of it being purple flowered (3/4). Use that logic to complete the table below.

Probability of being Tall:

_______

Probability of having purple flowers:

__________

Probability of being tall, with purple flowers:

___________

Probability of being Tall:

_______

Probability of having white flowers:

__________

Probability of being tall, with white flowers:

___________

Probability of being short:

_______

Probability of having purple flowers:

__________

Probability of being short, with purple flowers:

___________

Probability of being short:

_______

Probability of having white flowers:

__________

Probability of being short, with white flowers:

___________

[l]1/4

[fx] No, that’s not correct. Please try again.

[f*] Correct!

[l]3/4

[fx] No, that’s not correct. Please try again.

[f*] Excellent!

[l]9/16

[fx] No. Please try again.

[f*] Correct!

[l]3/16

[fx] No, that’s not correct. Please try again.

[f*] Excellent!

[l]1/16

[fx] No, that’s not correct. Please try again.

[f*] Excellent!

[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 this 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 cell 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 homologue, and a recessive allele (t or p) on the other homologue.

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 in your student learning guide, 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.

[q labels = “right”]In the dihybrid cross we’ve been examining, each of the parents can produce four types of gametes: TP, Tp, tP and tP. That means that  to predict the genotypes and phenotypes of the offspring, you need a 4 x 4 square to accommodate all the possible zygotes that can be produced.

I’ve set up the square below. Your job is to drag the right genotypes into the right spot.

TP Tp tP tp
TP _______ _______ _______ _______
Tp _______ _______ _______ _______
tP _______ _______ _______ _______
tp _______ _______ _______ _______

[l]TTPP

[fx] No. Please try again.

[f*] Great!

[l]TTPp

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]TtPP

[fx] No, that’s not correct. Please try again.

[f*] Correct!

[l]TtPp

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]TTpp

[fx] No, that’s not correct. Please try again.

[f*] Excellent!

[l]Ttpp

[fx] No. Please try again.

[f*] Correct!

[l]ttPP

[fx] No. Please try again.

[f*] Good!

[l]ttPp

[fx] No. Please try again.

[f*] Excellent!

[l]ttpp

[fx] No. Please try again.

[f*] Good!

 

[q labels = “right”] Here’s the Punnett square you completed on the previous slide. Let’s analyze the results to make sure we’re still seeing that 9:3:3:1 ratio.

  • Drag a ♠ if the phenotype is Tall stem, Purple Flowers
  • Drag a ♣ for Tall, white flowered
  • Drag a ♥ for short, purple flowered
  • Drag a ♦ for short, white flowered
TP Tp tP tp
TP TTPP __ TTPp __ TtPP __ TtPp __
Tp TTPp __ TTpp __ TtPp __ Ttpp __
tP TtPP __ TtPp __ ttPP __ ttPp __
tp TtPp  __ Ttpp __ ttPp __ ttpp __

[l]

[fx] No, that’s not correct. Please try again.

[f*] Correct!

[l]

[fx] No. Please try again.

[f*] Great!

[l]

[fx] No. Please try again.

[f*] Correct!

[l]♦

[fx] No, that’s not correct. Please try again.

[f*] Good!

[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].

[c]nine

[c]three

[c]three

[c]one

[/qwiz]

3. Solving Dihybrid Cross (and related) Problems

Mendel’s principle of independent assortment is important to know about in its own right. It’s also the basis of many dihybrid cross genetics problems that you’ll be asked to solve in a typical introductory college-level/AP biology course.

In solving dihybrid cross problems, the toughest part, for most students, seems to be figuring out the gametes that the parents can produce. Here’s an example, working with Mendel’s peas.

Try to solve these problems on your own before looking at the answer.

SAMPLE PROBLEM 1

In peas, round seeds (R) dominates over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). What is the genotype of an organism that is heterozygous round and heterozygous yellow? What gametes could this organism form?

Step 1: Figure out the genotype of the parents

The parent is described as heterozygous round and heterozygous yellow. That means it’ll be RrYy

Step 2: Use the FOIL algorithm to determine what gametes this parent can make.

FOIL (which I mentioned in the interactive reading above) stands for First, Outside, Inside, Last.

  • “First” means the first allele in each pair. “R” is the first allele in “Rr,” and “Y” is the first allele in “Yy.” So, the first alleles in the genotype RrYy are RY. That’s the first possible gamete.
  • Outside means the two outermost alleles. In RrYy that’s Ry. That’s the second possible gamete.
  • Inside means the two innermost alleles. In RrYy that’s rY. 
  • Last means the two last alleles in each pair. In RrYy that’s ry

So, the four possible gametes are RY, Ry, rY, and ry.

SAMPLE PROBLEM 2

The set up is the same as in problem 1: In peas, round seeds (R) dominates over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). What is the genotype of an organism that is homozygous round and heterozygous yellow? What gametes could this organism form?

Step 1: Figure out the genotype of the parents

The parent is described as homozygous round and heterozygous yellow. That means it’ll be RRYy.

Step 2: Use the FOIL algorithm to determine what gametes this parent can make.

  • “First” means the first allele in each pair. So, the first alleles in the genotype RRYy are RY. That’s the first possible gamete.
  • Outside means the two outermost alleles. In RRYy that’s Ry. That’s the second possible gamete.
  • Inside means the two innermost alleles. In RRYy that’s RY. 
  • Last means the two last alleles in each pair. In RRYy that’s Ry

If you list the four alleles derived using FOIL, you’ll see RY, Ry, RY, and Ry. The last two are repeats, which means that this parent can produce only two types of gametes: RY and Ry

SAMPLE PROBLEM 3

Let’s put these two problems together.

In peas, round seeds (R) dominates over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). What is the result of a cross between a dihybrid Round Wrinkled parent and a parent that is homozygous round and heterozygous yellow?

STEP 1: Figure out the genotype of the parents. Look at problem 1 and 2 above. The genotypes are RrYy x RRYy.

STEP 2: Figure out the genotype of the gametes each parent can make.

Look at problems 1 and 2 above for the solution.

RrYy: RY, Ry, rY, ry

RRYy: RY and Ry.

STEP 3: Set up a Punnett Square.

In introducing dihybrid crosses, we used a Punnett square that was 4 squares x 4 squares, to accomodate all the possible gametes. But in the cross above, one parent can only form two types of gametes. So you can set up your Punnett square in a 4 x 2 grid, like this:

RY Ry rY ry
RY
Ry

STEP 4: Complete your Punnett square by combining the alleles in the gametes to see the possible offspring.

RY Ry rY ry
RY RRYY RRYy RrYY RrYy
Ry RRYy RRyy RrYy Rryy

STEP 5: Analyze the offspring.

When we were doing monohybrid crosses, I had you list both the genotypes and phenotypes of the offspring. With dihybrid crosses, you mostly have to report phenotypes, and their frequency.

In this case, the analysis is simplified because none of the offspring are wrinkled, so there are only two phenotype categories. I’m going to put a ♥ in all the squares with offspring that are round and yellow (R_Y_) and a ♦ in all the squares that have offspring that are round and green (R_yy).

RY Ry rY ry
RY RRYY♥ RRYy♥ RrYY♥ RrYy♥
Ry RRYy♥ RRyy♦ RrYy♥ Rryy♦

75% of the offspring are round and yellow

25% of the offspring are round and green.

4. Practice Problems

Solve the following problems in your student learning guide. Then flip the card to check your answer.

[qdeck qrecord_id=”sciencemusicvideosMeister1961-Dihybrid Crosses, Practice Problems”]

[h]Dihybrid Cross Problems (Problems 1 and 2 are sample problems).

[i]Biohaiku

Dihybrid Crosses

Independent Assortment

Genes flowing through time

[q]Problem 3: In peas, round seeds (R) dominates over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). What is the genotype of an organism who is homozygous wrinkled and homozygous yellow? What would be the possible genotypes of its gametes?

[a]The genotype of the homozygous wrinkled and homozygous yellow parent is rrYY. It can only produce one type of gamete: rY.

[q]Problem 4: In peas, round seeds (R) dominates over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). Cross a homozygous round, heterozygous yellow plant with one that is heterozygous round and homozygous green. What are the phenotypes of the offspring?

[a]Step 1: Genotypes of the parents: RRYy x Rryy

Step 2: Genotypes of the parents gametes:  RRYy can create gametes RY and Ry. Rryy can create gametes Ry and ry.

Steps 3 and 4: Punnett square

Ry Ry
Ry Rryy Rryy
ry Rryy Rryy

STEP 5: Results

1 round and yellow: 1 round and green

[q]Problem 5: In watermelon, green (G) is dominant to striped (g). Short (S) is dominant to long (s). What is the genotype of an organism that is heterozygous green and heterozygous short?

[a]Heterozygous green and heterozygous short is genotype GgSs

[q]Problem 6: In watermelon, green (G) is dominant to striped (g). Short (S) is dominant to long (s). Create a Punnett square showing a dihybrid cross, and list the frequency of each phenotype.

[a]STEP 1: Parental genotypes are GgSs x GgSs

STEP 2: Both parents will produce gametes GS, Gs, gS, and gs.

STEPS 3 and 4: Punnett Square

GS Gs gS gs
GS GGSS GGSs GgSS GgSs
Gs GGSs GGss GgSs Ggss
gS GgSS GgSs ggSS ggSs
gs GgSs Ggss ggSs ggss

STEP 5: 9 Green short, 3 green long, 3 striped short, 1 striped long

[q]Problem 7: In humans, the gene for normal skin color (A) dominates the gene for albino skin (a). The gene for normal body height (M) dominates the gene for being a dwarf (m). Both genes are autosomal. A heterozygous normal skin and heterozygous normal height man has children with a heterozygous normal skin dwarf woman. What are the resulting phenotypes?

[a]STEP 1: Parental Genotypes: AaMm x Aamm

STEP 2: Genotypes of gametes: AM, Am, aM, am for parent 1 ; Am and am for parent 2

STEPS 3 and 4: Punnett Square

AM Am aM am
Am AAMm AAmm AaMm Aamm
am AaMm Aamm aaMm aamm

STEP 5: 3 Normal skin, normal height; 3 normal skin dwarfs; 1 albino, normal height, 1 albino dwarf.

[q]Problem 8: In humans, the gene for normal skin color (A) dominates the gene for albino skin (a). The gene for normal body height (M) dominates the gene for being a dwarf (m). A homozygous normal skinned dwarf has children with an albino dwarf. What are the resulting phenotypes?

[a] STEP 1: Parental Genotypes: AAmm x aamm

STEP 2: Genotypes of gametes: Am for parent 1 ; am for parent 2

STEPS 3 and 4: Punnett Square

Am
am AAMm

STEP 5: 100% of offspring are normal skin dwarfs.

[/qdeck]

Next Steps

  1. Continue on to Linked Genes (the next tutorial in this series on genetics)
  2. Return to the Genetics Main Menu