Introduction and Goals
This tutorial will teach you how to predict the segregation of alleles in the formation of gametes by parents that are heterozygous for different characters. Initially you will work with a tool referred to as a Punnett square, but later you will see how determining probabilities can help you make the same predictions much more easily. By the end of this tutorial you should have a working understanding of:
- The application of Punnett squares for monohybrid and dihybrid crosses
- Why test crosses are used to determine some genotypes
- Basic probability theory
- When to apply the Rule of Multiplication and the Rule of Addition
- Diagram how to use a Punnett square to determine the expected genotypes and phenotypes for a monohybrid cross
- Explain the application of a test cross to determine the genotype
- Discuss why it becomes more challenging to use a Punnett square as the number of characters increases
- Solve genetics problems using the Rule of Multiplication and the Rule of Addition to determine the probability of expected genotypes and phenotypes
A monohybrid cross involves the crossing of individuals and the examination of a single character (flower color [Figure 1] OR seed color OR pod shape, etc.) in their offspring. The Punnett square is a useful tool for predicting the genotypes and phenotypes of offspring in a genetic cross involving Mendelian traits. Constructing a Punnett square is quite easy, as shown in the Web sites below.
Figure 1. Mendel's law of segregation. (Click image to enlarge)
Problem 1: The Monohybrid Cross - This tutorial teaches how to set up a Punnett square and how to interpret the results. After viewing the tutorial, close the Monohybrid Cross Problem Set window to return to this page. Provided by The Biology Project, from the University of Arizona. While we’re on the subject, The Biology Project is a great interactive resource for learning about biological topics as diverse as immunology and genetics. Check out their Web site at http://www.biology.arizona.edu/ for additional information, supplemental problems, and further enrichment.
Problem 3: Mendel's Experiment 1 - Find the correct answer to the multiple-choice, monohybrid cross question. Work out the problem using a piece of paper and pencil. To view an explanation of the problem, select the "TUTORIAL" button. After viewing the correct answer, close the Monohybrid Cross Problem Set window to return to this page. This problem is a part of the Monohybrid Problem Set provided by The Biology Project, from the University of Arizona.
This is an animated Punnett square diagram of a monohybrid cross.
Interpreting the Results of a Punnett Square
The key to understanding Punnett squares is in realizing that the expected types of offspring are probabilities. The Punnett square is useful because it calculates the probability of producing each of the genotypes or phenotypes for any one offspring. Let's look at an example. A self-pollination of F1 hybrids for flower color (Pp) gives a 3:1 ratio of purple:white flowers in the offspring. This means that 75% of the offspring are predicted to have purple flowers and 25% white flowers. It also means that for every offspring produced, that offspring has a 75% chance of having purple flowers and a 25% chance of having white flowers. Therefore, one would expect the number of offspring with a particular trait to be equal to the probability multiplied by n, the number of offspring. As you learned in the last tutorial when working with ratios, if you are given the ratio and the total number of offspring counted, you can convert the ratio to a probability and predict the number of offspring with each trait.
A test cross is used to determine the genotype of an individual with a dominant trait. Because the trait is dominant, an individual with the trait could be homozygous or heterozygous for the trait. (This cannot always be determined by simply looking at the phenotype of the individual.) In a test cross, an individual with the dominant phenotype is crossed with a fully recessive individual. As shown in Figure 2, there are two possible outcomes (depending on the genotype of the parent). By examining the offspring of the cross, the unknown genotype of the dominant parent can be determined.
Figure 2. A test cross. (Click image to enlarge)
Problem 7. The Test Cross - Find the correct answer to the multiple-choice test cross question. To view an explanation of the problem, select the "TUTORIAL" button. After viewing the correct answer, close the Monohybrid Cross Problem Set window to return to this page. Provided by The Biology Project, from the University of Arizona.
Constructing a Punnett square for a dihybrid cross is similar to the method used for a monohybrid cross. Determine the alleles produced by each parent, draw the Punnett square, and then combine the gametes for each cell. The following Web sites should help to demonstrate the process more clearly.
Problem 1: Predicting gametes in a dihybrid cross - This tutorial teaches how to predict combinations of alleles in gametes of plants that are heterozygous for two traits. After viewing the tutorial, close the Dihybrid Cross window to return to this page. Provided by The Biology Project, from the University of Arizona.
Problem 3: A genetic cross yielding a 9:3:3:1 ratio of offspring - This tutorial teaches how to set up a Punnett square for a dihybrid cross and how to interpret the results. After viewing the tutorial, close the Dihybrid Cross window to return to this page. Provided by The Biology Project, from the University of Arizona.
Problem 7: Homozygous offspring of a dihybrid cross - Find the correct answer to the multiple-choice, dihybrid cross question. You will likely find it helpful to pull out a piece of paper and pencil in order to solve this problem. To view an explanation of the problem, select the "TUTORIAL" button. After viewing the correct answer, close the Dihybrid Cross Problem Set window to return to this page.
This problem is a part of the Dihybrid Problem Set provided by The Biology Project, from the University of Arizona.
This is an animated Punnett square diagram of a dihybrid cross.
Meiosis: The Mechanism for Segregation of Alleles
Recall from Tutorial11, that homologous chromosomes pair during Prophase 1 of meiosis, and later, each member of the pair will be packaged into separate gametes. Therefore, as we discussed in Tutorial 28, alleles on homologous chromosomes will segregate during Meiosis I (The Law of Segregation), and if the genes for different characters reside on different chromosomes, they will also independently assort during Metaphase I. (The Law of Independent Assortment).
As the connection between genes and chromosomes started to become more clear in the years after Mendel's death, scientists noted what seemed to be a paradox in Mendel's Law of Independent Assortment and the behavior of chromosomes during gamete formation. Complex organisms must have huge numbers of genes to govern their biochemical processes, and yet, the number of chromosomes in these organisms is often small. For example, humans have 23 pairs of chromosomes, whereas pea plants have 7 pairs. Therefore, each chromosome must have many genes.
But if there are many genes on each chromosome, how can two genes on the same chromosome be separated from one another? The chromosome doesn't break apart during meiosis; therefore, there must be another explanation. The answer is that some genes are physically linked to one because they are located on the same chromosome, but they appear to independently assort because the chromosomes recombine (exchange genetic material) with one another. The frequency with which they do this enables scientists to "map" gene locations on chromosomes. We will examine recombination in greater detail in upcoming exercises.
Since the alleles (traits) for different characters segregate from one another, probability can be used to mathematically predict the outcomes of genetic crosses involving one or more characters. Although Punnett squares are an effective and straightforward way to examine simple genetic crosses, they are unwieldy and time consuming when the inheritance of multiple characters is examined. A more effective way to examine inheritance is to use two basic laws of probability (The Rule of Multiplication and The Rule of Addition) to determine the probabilities of different outcomes when two individuals are crossed. A little background on probability will help your understanding. Probability scales range from 0 to 1. An event that is certain to occur has a value of 1, whereas an event that is certain not to occur has a value of 0. When you flip a coin, the probability of getting heads is 1/2 or 0.50, and the probability for getting tails is also 1/2 or 0.50. If you are rolling a single die, the probability of rolling a 1 is 1/6 or 0.17, as is the probability of rolling any other number on the die.
If the outcome of a probability event is unaffected by what has happened in previous trials, it is an independent event. When you roll a die, the chance of rolling a 1 is always 1/6, regardless of what you rolled previously. A lottery drawing, on the other hand, is an example of a non-independent event. If balls numbered 1 through 36 are placed in the lottery machine and mixed, the odds of getting any of the numbers is 1/36 on the first number drawn. However, on the second number, the odds decline to 1/35 because the removal of the first ball has affected the odds on the second number. For our purposes, we will be working almost entirely with independent events; however, it is important to understand the distinction.
Rule of Multiplication
Figure 3 shows the role of probability when estimating offspring genotypes in Mendelian genetics. The chance of getting a particular offspring requires two independent events to occur: the segregation of alleles into ova in the female, and the segregation of alleles into sperm in the male. In this example, each of the parents is heterozygous for flower color (Pp); because the two alleles in each individual segregate randomly during meiosis, the chance of getting either a P or p allele in an ova or sperm is 50%. To form an offspring with a particular genotype, the appropriate gametes must come together. The probability of this happening is determined by multiplying the probabilities of the two independent events that must occur to give the desired outcome.
Figure 3. Allele segregation, fertilization, and probability. (Click image to enlarge)
In the example given, the chance of getting a homozygous dominant offspring (PP) would be equal to the chance of getting a P allele in the ova (50%) multiplied by the chance of getting a P allele in the sperm (50%). Much like flipping two coins in the hope of both coming up heads, the probability of getting an offspring of this genotype would be (1/2)(1/2) = 1/4. The same would be true of an offspring with a pp genotype, as the probability of getting two p alleles would also be (1/2) (1/2) = 1/4.
These examples demonstrate the Rule of Multiplication, which allows one to calculate the probability of two or more independent events occurring together in a specific combination. To determine the probability of an event occurring, determine the individual probabilities of each independent event, then multiply the individual probabilities to obtain the probability of these events occurring together. Some non-genetic examples should help to clarify this concept.
(a) What is the probability of getting heads three times in a row when flipping a coin?
- Probability of getting heads on one flip = 1/2
- Probability of getting heads on three consecutive flips = (1/2) (1/2)(1/2) = 1/8(b) What is the probability of rolling a six on two consecutive rolls of a die?* Probability of getting a six on one roll = 1/6
- Probability of getting a six on two consecutive rolls = (1/6)(1/6) = 1/36
Rule of Addition
The Rule of Multiplication works well when there is only one way to arrive at a particular outcome, but what if the outcome can occur in different ways? For example, what is the probability of getting an offspring with the genotype Pp in the cross shown in Figure 4? Notice that there are two ways to get this genotype: a P from the ova and a p from the sperm, and vice versa. To calculate this probability, use the Rule of Addition. This rule states that if an event can occur in more than one way, the probability of that event occurring is equal to the sum of the probabilities of each way the event can occur.
Figure 4. Allele segregation, fertilization, and probability. (Click image to enlarge)
As depicted here, the probability of getting an offspring with the genotype Pp is equal to the probability of getting Pp (P from the ova, p from the sperm) plus the probability of getting pP (p from the ova, P from the sperm). Therefore, the probability would be 1/4 + 1/4 = 2/4 = 1/2. This example demonstrates how both the Rule of Multiplication and the Rule of Addition can be used to predict the outcome of genetic crosses. To ensure that you understand the Rule of Addition, here is a non-genetic example.
What is the probability of getting heads, at least once, in two flips of a coin? There are three possible ways to do this: heads on both flips, heads on the first flip, or heads on the second flip.
(a) Use the Rule of Multiplication to calculate the probabilities of each event that satisfies the conditions of the question.
- Probability of getting heads on first flip, heads on second flip = (1/2)(1/2) = 1/4
- Probability of getting heads on first flip, tails on second flip = (1/2)(1/2) = 1/4
- Probability of getting tails on first flip, heads on second flip = (1/2)(1/2) = 1/4
(b) Use the Rule of Addition to calculate the overall probability.
- Probability of getting heads, at least once, in two flips of a coin = 1/4 (heads on the first flip, tails second) + 1/4 (tails first, heads second) + 1/4 (heads both times)= 3/4
The phenotype of an organism arises from the complex interactions of many character traits. In a monohybrid cross, the inheritance of one character is predicted. A testcross allows the determination of the genotype of a parent with a completely dominant phenotype (homozygous dominant and heterozygotes look the same). In a dihybrid cross two characters can segregate in several ways, with each leading to a different phenotype. How can one predict the likelihood of the different phenotypes from a dihybrid cross? This becomes relatively straightforward, providing each character trait behaves in a Mendelian fashion; that is, a simple dominant-recessive expression pattern. If the character traits are located on different chromosomes, they will segregate independently from each other. This means that the separate probabilities of each character trait can be predicted, as with a monohybrid cross. It follows then that the probabilities of these different phenotypes occurring simultaneously in the same plant will be the product of their probabilities occurring alone. Using the Rules of Multiplication and Addition will allow you to solve genetics problems without having to use a Punnett square, which becomes cumbersome even for a dihybrid cross.
In other words, if a purple/round seeded plant that is heterozygous for both characters is crossed with itself ("selfing"), then the probability of observing round/seeded progeny will be 0.75 x 0.75 = 0.56. Be sure that you understand the logic behind this prediction and how it was calculated. Also, if there are more than two ways a given genotype can arise, then the probability is the sum of each way it can arise.
Relatively few character traits are inherited in a simple Mendelian manner. There are a number of other basic patterns of inheritance that have been described, including incomplete dominance, codominance, multiple alleles, pleiotropy, epistasis and polygenic inheritance. These will be discussed in the next tutorial.
After reading this tutorial, you should have a working knowledge of the following terms:
- dihybrid cross
- monohybrid cross
- Punnett square
- Rule of Addition
- Rule of Multiplication
- test cross
Case Study for Predicting Phenotypes and Genotypes
Your best friend is from a family that has 5 siblings – 4 boys and 1 girl. She also plans on having 5 children when she starts a family.
- What is the probability that she will have 4 boys and 1 girl?
- Is the probability that you calculated above the same as the probability that her first 4 children will be boys and her 5th child will be a girl?
- Why are these probabilities the same or different?
Now that you have read this tutorial and worked through the case study, go to ANGEL and complete the tutorial practice problems to test your understanding. Questions? Either send your instructor a message through ANGEL or attend an online office hour (the times are posted on ANGEL).