CONCEPTS IN BIOLOGY

PART III. MOLECULAR BIOLOGY, CELL DIVISION, AND GENETICS

 

10. Patterns of Inheritance

 

10.5. Solving Genetics Problems

 

Many students become confused when they try to solve genetics problems because they are not sure where to begin or when it is appropriate to apply a principle. As a result, they move directly to drawing a Punnett square and begin to incorrectly fill it with letters. Developing an organized and consistent strategy will help solve such problems.

 

Single-Factor Crosses

Problem Type: Single-Factor Cross

INTRODUCTORY PRINCIPLES: Genetics problems can vary greatly in complexity and in the type of information that is provided. Let’s start with a genetics problem that considers a single trait for which there are two alleles.

Cross 1: The pod color of some pea plants is inherited so that green pods are dominant to yellow pods. A pea plant that is heterozygous for green pods is crossed to a pea plant that produces yellow pods. What proportion of the offspring will have green pods?

 

Gene Key

Gene or Condition: pod color

Allele

Possible

 

Symbols

Genotypes

Phenot

G = green

GG

Green

 

Gg

Green

g = yellow

gg

Yellow

 

The question describes a gene affecting pea pod color. The two different phenotypes are green and yellow. The question also states that “green pods are dominant to yellow pods.” Because of this statement, use a capital letter to represent the green allele and a lowercase letter to represent the yellow allele. We are using the letter G, but any other letter would work. The only requirement is to use the same letter for both alleles. The gene key table shows the type of information needed to describe how the alleles for a characteristic work together. A gene key is a reference that will help solve the problem.

Now, organize the actual genetic cross. Table 10.2 shows each step in a simple genetics problem and skills that may be necessary to move from one step to the next. This problem starts at the top row and works toward the bottom of the table. There are several steps in this process. The steps are represented by each row in the table. The rows are titled. “Parental phenotypes,” “Parental genotypes,” “Possible sex cells,” “Offspring genotype,” and “Offspring phenotype.” First, determine what information is provided in the question about the organisms that are involved in this cross. Identify the following pieces of information in the question.

• A pea plant with green pods is crossed to a pea plant with yellow pods.

• The green pea plant is heterozygous.

• The yellow pea plant is homozygous.

 

TABLE 10.2. Steps in Solving a Genetics Problem

Solution Pathway

 

The question we are trying to answer is, “What proportion of the offspring will have green pods?” Solve the problem by using the table as a guide. These steps describe the process:

1. The first statement about the organisms being crossed is shown in the “Parental phenotypes” row as Green x Yellow. Using this information, the gene key, and the remaining two statements, we determine the parental genotypes. The reasoning described in this process is in the “Parental genotypes” row of table 10.2.

2. The next step is to use the parental genotypes to determine the parents’ possible sex cells. It is necessary to apply Mendel’s Law of Segregation to do this correctly. This process is described in the “Possible sex cells” row.

3. Now that the parents’ gametes are identified, use a Punnett Square to predict the genotypes of the offspring. Create a Punnett square so that there is one row or column for each gamete. This process is shown in the “Offspring genotype” row.

4. Return to the gene key to determine the offspring phenotypes from the genotypes just produced with the Punnett square.

5. Finally, remember to look at the question that was asked. In this example, the question is what proportion of the offspring will produce green pea pods. The answer is 50%.

 

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In this example, all the information from the problem fits into the gene key and the first rows of table 10.2. Not all problems are like this. Sometimes, a problem provides information about the offspring and requests information about the parents. Table 10.2 will help you do this as well. The principles that applied as you worked down the table still apply as you go the other direction.

 

Problem Type: Single-Factor Cross

INTRODUCTORY PRINCIPLES: When both parents are heterozygous and the alleles are completely dominant and recessive to each other, the predicted offspring ratio is always 3:1 (75% of the dominant phenotype to 25% of the recessive phenotype.) If the genotypes of parents are not known and the offspring have a 3:1 ratio, then geneticists frequently infer that the parents are both heterozygous for the trait being considered.

To illustrate this 3:1 pattern of inheritance, consider the disorder phenylketonuria (PKU). People with phenylketonuria are unable to convert the amino acid phenylalanine into the amino acid tyrosine. The buildup of phenylalanine in the body prevents the normal development of the nervous system. Such individuals may become mentally retarded if their disease is not controlled.

Figure 10.4 shows the metabolic pathway in which the amino acid phenylalanine is converted to the amino acid tyrosine by the enzyme phenylalanine hydroxylase. Tyrosine is then used as a substrate by other enzymes. In the abnormal pathway, the substrate phenylalanine builds up, because the enzyme phenylalanine hydroxylase does not function correctly in people with PKU. As phenylalanine levels rise, it is converted to phenylpyruvic acid, which kills nerve cells.

 

 

FIGURE 10.4. Phenylketonuria (PKU)

PKU is a recessive disorder located on chromosome 12. The diagram on the left shows how the normal pathway works. The diagram on the right shows an abnormal pathway. If the enzyme phenylalanine hydroxylase is not produced because of a mutated gene, the amino acid phenylalanine cannot be converted to tyrosine and is converted into phenylpyruvic acid, which accumulates in body fluids. The buildup of phenylpyruvic acid causes the death of nerve cells and ultimately results in mental retardation. Because phenylalanine is not converted to tyrosine, subsequent reactions in the pathway are also affected.

 

Cross 2: The normal condition is to convert phenylalanine to tyrosine. It is dominant over the condition for PKU. If both parents are heterozygous for PKU, what is the probability that they will have a child who is normal? A child with PKU?

As in the previous example, use the gene key to summarize this problem:

• There are 2 alleles. One is responsible for the normal condition and the other is for PKU.

• The normal condition is dominant over PKU. From this statement, infer that individuals with a normal phenotype can be either PP or Pp.

 

Gene Key

Gene or Condition: Phenylketonuria

Allele

Symbols

Possible

Genotypes

Phenotype

P = normal

PP

Normal

 

Pp

Normal

p = phenylketonuria

pp

Phenylketonuria

 

1. The problem states that “both parents are heterozygous for PKU.” This describes the parental genotypes and determines the genotypes for both parents to be Pp. Enter this information on the “Parental Genotypes” row. Although it is not necessary to solve the problem, you can use the gene key to determine the parental phenotypes. Pp individuals are normal. Note: This genetic problem does not start with the first row of table 10.3. A genetics problem can start at any point in the table and require that you determine things about either the parents or the offspring.

2. Determine the possible sex cells. Pp individuals will have gametes that are P and p.

3. Determine the offspring genotypes using a Punnett square. Create your Punnett square so that there is one row or column for each gamete. Note that, in this situation, three different genotypes are produced—PP, Pp, and pp.

4. Determine the offspring phenotypes by using the gene key and combining genotypes with similar phenotypes.

5. Finally, answer the question that was asked from the problem. In this case, two questions were asked. The probability of having a normal child is 75%. The probability of having a child with PKU is 25%.

 

TABLE 10.3. Solution Pathway

 

Double-Factor Crosses

Up to this point, we have worked only with single-factor crosses. Now we will consider how to handle genetics problems that involve following two distinct characteristics—double-factor crosses. In solving double-factor crosses, it is important to consider the principle Mendel identified as the Law of Independent Assortment. The Law of Independent Assortment states that alleles of one characteristic separate independently of the alleles of another. This law is applied only when working with two genes for different characteristics that are on different chromosomes. This is an important distinction, because genes that are positioned near each other on a chromosome tend to stay together during meiosis and therefore tend to be inherited together. If genes are inherited together, they are not assorting in a random manner. Their assortment is not independent of each other.

In genetics problems, the process of predicting the sex cells that can be produced in double-factor crosses is affected by independent assortment. The following example illustrates how independent assortment works. Recall that if an individual has the genotype Aa, we predict that 50% of his or her reproductive cells have the A allele and 50% have the a allele. This is an application of the Law of Segregation. If an individual has the genotype Bb, we can make a similar prediction with regard to the B and b alleles. What happens when we want to look simultaneously at both sets of alleles when the “A” characteristic is on a different chromosome from the “B” characteristic? What are the possible sex cells that could be produced for an individual that is AaBb?

In order to answer this question, we have to apply both the Law of Segregation and the Law of Independent Assortment. For now, we will assume that the two genes are on different chromosomes. As mentioned earlier, the law of segregation predicts that 50% of the gametes will have A and 50% will have a. Likewise 50% of the gametes will have B and 50% b. The Law of Independent Assortment says that, if a gamete receives an A allele, it has an equal chance of also receiving a B allele or a b allele. Thus, the sex cells that are predicted are AB, Ab, aB, and ab. Notice that

• every sex cell has either an A or an a but not both. Every sex cell has either a B or a b but not both. This means that all the sex cells have 1 and only 1 of the 2 alleles for each characteristic.

• each allele is found in 50% of the sex cells.

• the alleles for one characteristic are inherited independently from the other.

You can check to see if you have made correct choices by making sure that only one allele for each characteristic is present in a sex cell. Note that sex cells with allele combinations such as AA or AaB are incorrect. Both of these incorrect examples have more than 1 allele for a gene. The first example (AA) should have only a single A and is missing a copy of the B gene altogether. The second example (AaB) should have either the A or the a allele, but not both.

 

Problem Type: Double-Factor Cross

Cross 3: In humans, the allele for free earlobes is dominant over the allele for attached earlobes. The allele for dark hair dominates the allele for light hair. If both parents are heterozygous for earlobe shape and hair color, what types of offspring can they produce, and what is the probability for each type?

Just as in a single-factor cross, start by creating a gene key. You are working with two characteristics this time, so create a key for both. Remember that not all the information in the gene key is stated directly in the problem. From the problem, you should be able to identify that

• There are two genes—earlobe type and hair color.

• The free earlobe allele is dominant to the attached earlobe allele.

• The dark hair allele is dominant to the light hair allele. From this information, you should be able to infer that

• Because the free earlobe allele is dominant, it can have two genotypes—EE and Ee.

• Because dark hair is dominant, it can have two genotypes—HH and Hh.

 

Gene Key

Gene or Condition: earlobe type

Allele

Possible

 

Symbols

Genotypes

Phenotype

E = free

EE

Free earlobes

 

Ee

Free earlobes

e = attached

ee

Attached earlobes

 

Gene or Condition: hair color

Allele

Possible

 

Symbols

Genotypes

Phenotype

H = dark hair

HH

Dark hair

 

Hh

Dark hair

h = light hair

hh

Light hair

 

1. After you have the gene key complete, move on to the cross setup in table 10.4. The problem states that “both parents are heterozygous for earlobe shape and hair color.” This is a description of the parents’ genotypes. It means that both parents are EeHh. Place this information on the “Parental Genotypes” row. Although it is not necessary, you can use the gene key to determine what the parent’s phenotypes are for earlobe type and hair color.

2. Determine the possible sex cells. EeHh individuals will have gametes that are EH, Eh, eH, and eh. This answer uses the Law of Segregation and the Law of Independent Assortment.

3. Determine the offspring genotypes using a Punnett square. Create your Punnett square so that there is one row or column for each gamete. Your Punnett square will create a 4 x 4 grid. Fill in the genotypes as shown.

4. Determine the offspring phenotypes by using the gene key and combining genotypes with similar phenotypes. In this problem, there will be four different groupings: (a) free earlobes and dark hair, (b) free earlobes and light hair, (c) attached earlobes and dark hair, and (d) attached earlobes and light hair.

5. Answer the question that was asked from the problem. The ratios are 9:3:3:1.

In cases where the alleles for each gene are completely dominant and recessive to each other and both parents are heterozygous for both characteristics, the predicted offspring ratio is 9:3:3:1. When scientists observe a 9:3:3:1 ratio, they suspect that both parents are heterozygous for both characteristics being considered.

 

TABLE 10.4. Solution Pathway

 

10.5. CONCEPT REVIEW

13. What does it mean when geneticists use the term independent assortment?

14. What is a Punnett square?

15. What is the probability of each of the following sets of parents producing the given genotypes in offspring?

Parents

Offspring

a. AA x aa

Aa

b. Aa x Aa

Aa

c. Aa x Aa

aa

d. AaBb x AaBB

AABB

e. AaBb x AaBB

AaBb

f. AaBb x AaBb

AABB

16. What possible combinations of parental genotypes could produce an offspring with the genotype Aa?

17. In certain pea plants, the allele T for tallness is dominant over t for shortness.

a. If a homozygous tall and homozygous short plant are crossed, what will be the phenotype and genotype of the offspring?

b. If both individuals are heterozygous, what will be the phenotypic and genotypic ratios of the offspring?

18. Certain kinds of cattle have two alleles for coat color: R = red, and r = white. when an cow is heterozygous, it is spotted with red and white (roan). When two red alleles are present, it is red. When two white alleles are present it is white. The allele L, for lack of horns, is dominant over l, for presence of horns. If a bull and a cow both have the genotype RrLl, how many possible phenotypes of offspring can they have?