SAT Biology E/M Subject Test

Part II: Subject Review

Chapter 8 Cracking Genetics

The SAT Biology E/M Subject Test will ask you some questions about genetics and inheritance. This chapter will explore the possible combinations of genes in an organism, and how these processes determine what an organism is like: namely, its traits.

BIOLOGY OF INHERITANCE

From studying earlier chapters, you already know that

•   genes direct the production of mRNA.

•   mRNA contains three-nucleotide-long segments called codons.

•   each codon codes for a single amino acid.

•   amino acid chains become proteins.

•   the primary function of proteins in a cell is to act as enzymes.

The enzymes inside an organism’s cells determine which chemical reactions will and will not occur, and that determines what the organism is like. So, an organism’s genes determine its features, its characteristics, its appearance—in other words, its traits. The genes determine what traits an organism will possess.

When talking about an organism’s traits, we say phenotype. If you’re a cat and your fur is gray, then we say that, in terms of fur color, your phenotype is gray. If your fur is white, then we say, in terms of fur color, your phenotype is white. If we meet a person who has diabetes, we say that, in terms of diabetes, his phenotype is diabetic. If we meet another person who does not have diabetes, we say that in terms of diabetes, her phenotype is nondiabetic, or we might say that her phenotype is normal. So phenotype just refers to the actual traits that an organism does and does not possess.

PHENOTYPE AND GENES

As you’ll remember, we said that, generally speaking, chromosomes are arranged in homologous pairs. Each member of the pair is similar to, but also different from, the other member.

Let’s look at one pair of chromosomes, and, in particular, let’s focus on one part of each chromosome. We want to make the picture easy to look at, so we won’t bother with actually shaping the chromosomes like a double helix/twisted ladder.

Notice that each of these shaded portions is labeled “black fur.” This means that the particular part of the chromosome that we’ve decided to look at codes for fur color. It’s responsible for producing the enzymes that catalyze the chemical reactions that determine the color of the organism’s fur. Now, this particular pair of chromosomes codes for enzymes that give rise to black fur. So, when it comes to fur color, what is this organism’s phenotype? Simple—it’s black.

Now, while we’re looking at these two chromosomes, let’s introduce another word: genotype. For the organism that we’re now talking about—the one with the black fur—we’d (1) look at the homologous chromosome pair pictured above, (2) see that the organism has black fur, and (3) say:

Fur Color Genotype: black/black (or BB)

Fur Color Phenotype: black


When we say phenotype, we’re talking about the organism and its traits. But when we say genotype, we’re talking about the genes responsible for those traits.


More About Genotype and Phenotype: Features That Are Dominant and Recessive

Let’s consider another organism with black fur. When it comes to fur, what’s the organism’s phenotype? It’s black.

Now let’s look at its genotype. Here are the chromosomes that contain the genes responsible for fur color.

Notice that the two homologous chromosomes don’t agree on what color the organism’s fur should be. One codes for black fur, and the other codes for white fur. Yet one member of the pair definitely wins the debate. The organism has black fur. When we want to describe this organism’s phenotype and genotype in terms of fur color we say:

Genotype: black/white (or Bb)

Phenotype: black

Why does one of the chromosomes get to express itself in the organism’s phenotype while the other one has to keep quiet? Here’s the answer: For the species to which this organism belongs, black fur is dominant, and white fur is recessive. That just means that if an organism of this species has one black-fur chromosome and one white-fur chromosome, the organism will have fur that’s black. Black fur is dominant—a chromosome that codes for black fur dominates a chromosome that codes for white fur. It gets to decide what the phenotype will be. White fur is recessive—it has to recede—it isn’t expressed if one of the pair of homologous chromosomes codes for black fur.

Notice, then, that in this example there are two “versions” of the gene for fur color—a black fur version and a white fur version. Versions of a gene are called alleles, and we can refer to the black fur allele or the white fur allele. In this example, the black fur allele is dominant over the white fur allele.

Notice that we can use a sort of “shorthand” to indicate genotype. We can use letters to represent the alleles an organism has. For example, in our first organism, the one in which both chromosomes code for black fur, we described the genotype as black/black, or BB. The uppercase B represents the allele for black fur. In our second organism, in which one chromosome codes for black fur and one codes for white fur, we described the genotype as black/white, or Bb. The lowercase b represents the allele for white fur.


Uppercase letters are used to represent the dominant allele, and lowercase letters are used to represent the recessive allele.


Notice also that the letter chosen to represent the dominant allele is the same letter chosen to represent the recessive allele. In other words, we didn’t switch letters and represent the allele for white fur with a lowercase w. As long as we’re talking about the same trait (in this case, fur color), the letter stays the same; we just use uppercase for dominant and lower case for recessive.

Let’s consider a few more organisms of this species. Here’s one with fur color genotype: white/white.

Its fur-color phenotype will be white. Even though white is recessive, both chromosomes agree that the fur should be white. There’s no black-fur chromosome around to dominate the matter.

As a matter of fact, if you see an organism of this species and its fur is white, you know that its genotype is white/white, or bb. Because white is recessive, that’s the only genotype that can produce a white phenotype. If a black-fur chromosome were around, the phenotype would have to be black. A white-fur phenotype definitely means a bb genotype because white is recessive.

Here’s an organism with fur-color genotype: white/black (bB).

Its fur-color phenotype will most definitely be black. Why? Because black fur is dominant and white fur is recessive. The black-fur chromosome dominates the white-fur one.

Time Out! Let’s Learn Some Simple Biology Subject Test Terms: Allele, Homozygous and Heterozygous, and Codominance

1. Allele

As we’ve already stated, the term allele is used to refer to a gene that gives rise to more than one version of the same trait—like, for example, eye color. Just now, for instance, we’ve been talking about the allele responsible for fur color. The black-fur and the white-fur genes are different alleles of the fur-color gene.

2. Homozygous and Heterozygous

When, for a particular trait, an organism’s two alleles are in agreement, we say the organism is homozygous for that trait. The organism that has genotype black/black (or BB) is homozygous for fur color. The organism that has genotype white/white (or bb) is also homozygous for fur color. In both cases the two alleles agree on fur color. On the other hand, organisms that had genotypes black/white (Bb) and white/black (bB) were cases in which there was a disagreement within the genotype. One allele codes for white, and the other codes for black. We say that these organisms are heterozygous for fur color, which means that their two alleles don’t agree on what color the fur should be. (We know, of course, that in both cases the fur will be black because, for this organism, black fur is dominant and white fur is recessive.)

3. Codominance

It is possible, in the case of multiple alleles (i.e., more than two), for two alleles to exhibit codominance. This means that a heterozygote expresses a mixture of the traits of both alleles. This is the case, for example, when someone has the blood type AB.

Quick Quiz #1

Fill in the blanks and check the appropriate boxes:

  1. Which term refers to an organism’s observable traits?

(A)  Phenotype

(B)  Genotype

(C)  Allele

(D)  Homozygous

(E)  Heterozygous

  2. The term allele [  is  is not] precisely synonymous with the term gene.

  3. Which of the following terms refers to the fact that the gene responsible for a particular observable trait might exist in more than one version?

(A)  Phenotype

(B)  Genotype

(C)  Allele

(D)  Homozygous

(E)  Heterozygous

  4. If, in a particular organism, one allele on one member of a homologous chromosome pair codes for blue eye color and a corresponding allele on the other codes for brown eye color, the organism is said to be [  homozygous  heterozygous] for eye color.

  5. If an organism is heterozygous for eye color, with one allele coding for green and the other allele coding for gray, the organism will have [  green eyes  gray eyes] if green is dominant, and [  green eyes  gray eyes] if gray is dominant.

  6. If, for a particular species, the allele that produces a disease called erythema is dominant and the corresponding allele that produces the absence of disease (a normal organism) is recessive, then

(A)  an organism with a genotype that is heterozygous for the trait will have the phenotype: [  normal  erythema].

(B)  an organism with a genotype that is homozygous for the dominant allele will have the phenotype: [  normal  erythema].

(C)  an organism with a genotype that is homozygous for the recessive allele will have the phenotype: [  normal  erythema].

(D)  an organism with a phenotype that is normal must have a genotype that is [  homozygous  heterozygous].

Correct answers can be found in Chapter 15.

Mating and Crossing: Predicting the Phenotype and Genotype of Offspring

When you take the SAT Biology E/M Subject Test you’ll be asked, in one way or another, to figure out what kind of genotypes and phenotypes to expect from a cross between two organisms that bear offspring. Suppose two organisms decide to mate and have offspring. One has black fur and the other has white fur. Here are the genotypes:

Male Parent

Female Parent

Bb

bb

Go back to Chapter 7 for a few minutes and reread the information about meiosis and fertilization. Realize that, from each pair of chromosomes normally present within the parent cell, the offspring gets one member from its father and another member from its mother. That’s because the spermatozoa and ova are haploid cells, each of which contributes one member of a homologous pair to the zygote, which, therefore, is diploid.

Realize also that any offspring gets its particular combination of chromosomes at random, and different offspring get different combinations. Here’s what we mean by that. For the chromosome that carries the fur-color allele, there are four possible combinations of homologous chromosomes that a particular offspring might receive from its parents.

1.   One animal, for instance, might get this from his mother and this one from his father:

That means the new organism will have genotype Bb. Its phenotype will be black, because black is dominant.

2.   Another possibility is that the animal might get this from his father and this from his mother:

This new organism will also have genotype Bb. Its phenotype for fur color will be black.

3.   In the third possible combination, an animal might get this from his father and this from his mother:

This new organism will have genotype bb. Its phenotype for fur color will be white.

4.   In the last possible combination, an animal might get this from his father and this from his mother:

That means the new organism will have genotype bb. Its phenotype for fur color will be white.

Take a Look at the Genotypes and Phenotypes of the Four Offspring These Parents Might Produce

If you look back at what we’ve just done, you’ll see that for these parents:

•   Two of the four possible offspring are genotype bb and phenotype white

and

•   Two of the four possible offspring are genotype Bb and phenotype black

We can say, therefore, that as for these parents and the fur color of their offspring, the probability is that:

•   50% of the offspring will be genotype bb and phenotype white

and

•   50% of the offspring will be genotype Bb and phenotype black

PUNNETT SQUARES

Before we go on with some more examples, let’s look at a quick and easy way to predict the possible genotypes and phenotypes of a given genetic cross. Draw a four-chambered box, and along the sides of the box indicate the genotypes of the two parents, like this:

Now, fill in the four segments of the box according to what you’ve drawn along the sides and you’ll easily visualize the four-genotype possibilities. (Knowing which trait is dominant and which is recessive means you can easily figure out the phenotype possibilities.) This is called a Punnett square.

From this, it’s quite easy to calculate the probability of each phenotype.

Once again, you come up with:

•   50% offspring of genotype bb and phenotype white

•   50% offspring of genotype Bb and phenotype black

We can say the probability of white fur is 50%. What can we say about the probability of black fur is also 50%.

Let’s Do Some More Predicting

Suppose some species of plant has the possibility of producing flowers that are (1) red or (2) yellow. Suppose we tell you, also, that yellow is dominant and red is recessive. Now let’s contemplate a cross between plants with these two genotypes:

Parent A

Parent B

Yy

Yy

Take a minute and draw a Punnett square in the space below. Fill it in as we did before. First we’ll ask you some questions, and then we’ll explain the answers so we’re sure you’re on top of this stuff.

Refer to your drawing and fill in the blanks and boxes appropriately.

•   Parent plant A has phenotype [  red flowers  yellow flowers].

•   Parent plant B has phenotype [  red flowers  yellow flowers].

•   Parent plants A and B [  can  cannot] possibly produce a plant with red flowers [  because  even though] both A and B have [  red flowers  yellow flowers].

•   The probability that parent plants A and B will produce a plant with red flowers is _____%.

•   The probability that parent plants A and B will produce a plant with yellow flowers is _____%.

Here’s How to Crack It

If you drew your Punnett square correctly, it probably looks something like this:

You might have drawn it with parent A on top and parent B along the side, or vice versa. It doesn’t matter. For each parent you might have arranged the y and Y in either order. That doesn’t matter either. You’d still get the same overall results.

To begin with, you know that each parent has phenotype yellow. Yellow is dominant. Even though each parent has a chromosome that wants the flowers to be red, each also has one that wants them to be yellow. Yellow wins the fight in both cases. Both parents have yellow flowers.

Your Punnett square shows you that if you make every possible combination, your four results are:

1.   a plant with genotype YY and phenotype yellow

2.   a plant with genotype Yy and phenotype yellow

3.   a plant with genotype Yy and phenotype yellow

4.   a plant with genotype yy and phenotype red

Of the four possible combinations, two give us genotypes that are heterozygous, which means the phenotypes are yellow, one gives us a genotype that’s homozygous for the yellow trait, which means that its phenotype is yellow, and one gives us a genotype that is homozygous for the red trait, which means that its phenotype is red.

Let’s Do Another One

Sticking with the red and yellow flowers, let’s draw some Punnett squares and make some predictions regarding the offspring of these two parent plants:

Parent A

Parent B

Yy

YY

Use the space below to draw this square as before. Make the appropriate markings and labels, and then fill in the blanks and boxes appropriately.

•   Parent plant A has phenotype [  red flowers  yellow flowers].

•   Parent plant B has phenotype [  red flowers  yellow flowers].

•   Parent plants A and B [  can  cannot] possibly produce a plant with red flowers.

•   The probability that parent plants A and B will produce a plant with red flowers is _____%.

•   The probability that parent plants A and B will produce a plant with yellow flowers is _____%.

Here’s How to Crack It

Your Punnett square should look like this:

One parent is homozygous for the dominant trait (yellow), and the other is heterozygous. For both parents, the phenotype is yellow. Your square shows you that the parents might possibly produce these four progeny:

1.   a plant with genotype Yy and phenotype yellow

2.   another plant with genotype Yy and phenotype yellow

3.   a plant with genotype YY and phenotype yellow

4.   another plant with genotype YY and phenotype yellow

The parents have a 50% probability of producing offspring with genotype Yy (or yY, however you want to look at it) and a 50% probability of producing offspring with genotype YY. In terms of phenotype, they have a 0% chance of producing offspring with red flowers and a 100% chance of producing offspring with yellow flowers. All of their offspring will have yellow flowers. Half will be heterozygous for the trait.

One More Example

Suppose a certain species of bird has the capacity to be born with oily or nonoily feathers. Suppose that oily is dominant and nonoily is recessive. Fill in the blank lines and boxes appropriately. (Hint: Figure out the genotypes of all organisms first.)

•   If an organism is born with nonoily feathers, it [  is  is not] possible that both of its parents were homozygous for oily feathers.

•   If an organism is born with oily feathers, it [  is  is not] possible that both of its parents were homozygous for oily feathers.

•   If an organism is born with oily feathers, it [  is  is not] possible that one of its parents was homozygous for nonoily feathers.

•   If an organism is born with nonoily feathers and its parent A had oily feathers, then

(a) the genotype for parent A was ____________,

(b) the genotype for parent B was __________, or __________,

(c) the phenotype for parent A was ____________,

(d) the phenotype for parent B was __________, or _________.

Here’s How to Crack It

•   If an organism is born with nonoily feathers, it [  is  is not] possible that both of its parents were homozygous for oily feathers.

If both parents are homozygous for oily feathers, your box would show you that all offspring would also be homozygous for oily feathers. That genotype can’t lead to nonoily feathers.

•   If an organism is born with oily feathers, it [  is  is not] possible that both of its parents were homozygous for oily feathers.

That’s because two parents that are homozygous for oily feathers will, in fact, produce offspring that are all oily-feathered. Your Punnett square will show you that.

•   If an organism is born with oily feathers, it [  is  is not] possible that one of its parents was homozygous for nonoily feathers.

That’s because an oily-feathered offspring can result from a parent that is homozygous for nonoily feathers so long as the other parent is either heterozygous or homozygous for oily feathers. Draw a Punnett square and you’ll see.

•   If an organism is born with nonoily feathers and parent A had oily feathers, then

(a) the genotype for parent A was Oo (or oO, if you want to look at it that way).

That’s because a nonoily-feathered organism can’t arise if either parent is homozygous for oily feathers. Because you’re told that parent A’s phenotype is oily, it’s got to be heterozygous: Oo.

(b) The genotype for parent B was Oo or oo.

That’s because a nonoily-feathered creature can result only if both parents have at least one chromosome for nonoily feathers. That means parent B has to be either Oo or oo.

(c) The phenotype for parent A was oily.

That’s because you’re told that parent A was oily-feathered.

(d) The phenotype for parent B was oily or nonoily.

That’s because a nonoily-feathered creature can result only if both parents have at least one chromosome for nonoily feathers. The right answer to item (b) tells you the right answer to this item (d). Parent B has genotype Oo or oo, which means its phenotype is either oily or nonoily.

Another Thing About Genetics and Inheritance: Sex and Sex-Linked Traits

Let’s think about people. As you know, their cells (other than sperm and ova) have 23 pairs of homologous chromosomes. One of those 23 pairs is the pair that determines whether a person is male or female, and this pair of chromosomes is known as the sex chromosomes. All other chromosomes are called autosomes.

The sex chromosomes have many genes that determine sexual attributes, as well as some genes that determine nonsexual attributes. A sex chromosome can either be male, in which case we call it a Y chromosome, or it can be female, in which case we call it an X chromosome. A person whose phenotype is male has the sex genotype XY, and a person whose phenotype is female has the sex genotype XX. The male got an X chromosome from his mother (whose genotype is XX) and a Y chromosome from his father (whose genotype is XY). The female got an X chromosome from her mother and an X chromosome from her father. So whether a child is born male or female depends on whether it gets an X or a Y chromosome from its father—because the father can donate either one.

MENDELIAN GENETICS

Gregor Mendel is known as the father is genetics. One of his hobbies was to study the effects of cross-breeding on different strains of pea plants. Mendel works exclusively with true-breeder pea plants. This means that the plants he used were genetically pure and consistently produced the same traits. For example, tall plants always produced tall plants; short plants always produced short plants. Through this work, he came up with three principles of genetics: the law of dominance, the law of segregation, and the law of independent assortment.

The Law of Dominance

Mendel crossed two true-breeding plants with contrasting traits: tall pea plants and short pea plants. This type of cross is called a monohybrid cross, which means that only one trait is being studied. In this case, the trait was height.

To his surprise, when Mendel mated these plants, the characteristics didn’t blend to produce plants of average height. Instead, all of the offspring were tall.

Mendel recognized that one trait must be masking the effect of the other trait. This is called the law of dominance. The dominant, tall allele, T, somehow masked the presence of the recessive, short allele, t. Consequently, all a plant needs is one tall allele to make it tall.

Monohybrid Cross

A simple way to represent a cross is to create a Punnett square. Let’s construct a Punnett square for the cross between Mendel’s tall and short pea plants. As we saw earlier, we can use the letter “T” for the tall, dominant allele and “t” for the short, recessive allele.

Since one parent was a pure, tall pea plant, we’ll give it two dominant alleles (TT homozygous dominant). The other parent was a pure, sort pea plant, so we’ll give it two recessive alleles (tt homozygous recessive). As you know, we should put the alleles for one of the parents across the top of the box and the alleles for the other parent across the side of the box.

Now we’ll fill in the four boxes by matching the letters. What are the results?

Each offspring received one allele from each parent. They all received one T and one t, so they are all Tt. Our parents had duplicate copies of single alleles, TT and tt, respectively. We could therefore refer to the parents as homozygous. The offspring, on the other hand, are heterozygous: They possess one copy of each allele.

Let’s compare the results of this cross with what we already know about meiosis. From meiosis, we know that when gametes are formed, the chromosomes separate so that each cell gets one copy of each chromosome. We now know that chromosomes are made up of genes, and genes consist of alleles. We’ve just seen that alleles also separate and recombine. We can say that, therefore, each allele in a Punnett square also “represents” a gamete:

When fertilization occurs, chromosomes—along with the alleles they carry—get paired up in a new combination.

The Law of Segregation

Next, Mendel took the offspring and self-pollinated them. Let’s use a Punnett square to spell out the results. This time we’re starting with the offspring of the first generation. Take a look at the results:

One of the offspring could be a short pea plant! The short-stemmed trait reappeared in the F2 generation (F1 would mean first generation, F2 means second generation). How could that happen again? Once again, the alleles separated and recombined, producing a new combination for this offspring. The cross resulted in one offspring with a pair of recessive alleles, tt. Because there is no T (dominant) allele around to mask the expression of the short, recessive allele, our new plant could wind up short.

Although all of the F1 plants appear to be tall, the alleles separate and recombine during the cross. This is an example of the law of segregation.

What about the genotype and phenotype for this cross? Remember, genotype refers to the genetic makeup of an organism, whereas phenotype refers to the appearance of the organism. Using the results of our Punnett square, what is the ratio of phenotypes and genotypes in the offspring?

Let’s sum up the results. We have four offspring with two different phenotypes: three of the offspring are tall, whereas one of them is short. On the other hand, we have three genotypes: 1 TT, 2 Tt, and 1 tt.

Here’s a summary of the results:

•   The ratio of phenotypes is 3 : 1 (three tall: one short)

•   The ratio of genotypes is 1 : 2 : 1 (one TT: two Tt: one tt).

The Law of Independent Assortment

So far, we have looked at only one trait: height (tall versus short). What happens when we study two traits at the same time? The two traits also segregate randomly. This is an example of independent assortment. For example, let’s look at two traits in pea plants: height and color. When it comes to height, a pea plant can be either tall or short. As for color, the plant can be either green or yellow, with green being dominant. This gives us four alleles. By the law of independent assortment, these four alleles can combine to give us four different gametes:

TG Tg tG tg

Dihybrid Cross

Keep in mind that the uppercase letter refers to the dominant allele. Therefore, “T” refers to tall and “G” refers to green, whereas “t” refers to short and “g” refers to yellow. Now let’s set up a cross between plants differing in two characteristics—called a dihybrid cross—using these four gametes and see what happens.

Each trait will act independently, meaning that a plant that is tall can be either green or yellow. Similarly, a green plant can be either tall or short. Here is the Punnett square for a cross between two double heterozygotes (Tt Gg):

This is an example of the law of independent assortment. Each of the traits segregated independently. Don’t worry about the different combinations in the cross—you’ll make yourself dizzy with all those letters. Simply memorize the phenotype ratio of the pea plants. For the 16 offspring there are:

•   9 tall and green

•   3 tall and yellow

•   3 short and green

•   1 short and yellow

That’s 9 : 3 : 3 : 1. Since Mendel’s laws hold true for most traits, that they might ask you about on the SAT Bio exam, simply learning the ratios of offspring for this type of cross will help you nail any questions that come up.

The Punnett square method works well for monohybrid crosses and helps us visualize the possible combinations. However, a better method for predicting the likelihood of certain results from a dihybrid cross is to apply the law of probability. For dihybrid ratios, the law states that the probability that two of more independent events will occur simultaneously is equal to the product of the probability that each will occur independently. To illustrate the product rule, let’s consider again the cross between two dihybrid tall, green plants with the genotype TtGg. To find the probability of having a tall, yellow plant, simply multiply the probabilities of each event. If the probability of being tall is  and the probability of being yellow is , then the probability of being tall and yellow is  ×  = .

One more thing: Probability can be expressed as a fraction, percentage, or decimal. Remember that this rule works only if the results of one cross are not affected by the results of another cross. Let’s review Mendel’s three laws:

Test Cross

Suppose we want to know if a tall plant is homozygous (TT) or a heterozygous (Tt). Its physical appearance doesn’t necessarily tell us about its genetic makeup. The only way to determine its genotype is to cross the plant with a recessive, short plant, tt. This is known as a test cross. Using the recessive plant, there are only two possibilities: (1) TT × tt or (2) Tt × tt. Let’s take a look:

If none of the offspring is short, our original plant must have been homozygous (TT). If, however, even one short plant appears in the bunch, we know that our original pea plant was heterozygous (Tt). In other words, it wasn’t a pure- breeding plant. A test cross uses a recessive organism to determine the genotype of an organism of unknown genotype.

Quick Quiz #2

Fill in the blanks and check the appropriate boxes:

  1. A male person receives from his father a(n) [  X  Y ] chromosome.

  2. A male person [  may  must ] have the genotype [  XY  XX  YY ].

  3. A female person [  may  must ] have the genotype [  XY  XX  YY ].

  4. In terms of sex, all persons, male or female, receive from their mothers a(n) __________ chromosome.

  5. In terms of sex, all females receive from their fathers a(n) _________ chromosome.

Correct answers can be found in Chapter 15.

Sex-Linked Traits

The sex chromosomes X and Y, as we said, carry certain genes on them. A sex-linked trait, is a trait whose allele is carried on one of the sex chromosomes. Almost all sex-linked traits have alleles that are carried on the X chromosome. The X chromosome is much larger than the Y chromosome and therefore carries more genes. For the purposes of the SAT Biology E/M Subject Test, the only sex-linked traits you need to worry about are ones carried on the X. You can, and should, think of them as being X-linked. The fact that some traits are X-linked leads to a few interesting phenotypes, which are very likely to show up on the exam. The three X-linked traits to know for the exam are hemophilia (a disorder of blood clotting), color blindness, and male pattern baldness.

Let’s consider the disease hemophilia, an X-linked trait. In other words, if a person carries the trait, he or she carries it on the X chromosome. It’s also recessive, so we can describe hemophilia as being an X-linked recessive trait. What does that mean? It means that a female can only get hemophilia if she’s homozygous for the X chromosome that carries the allele for hemophilia. For example, take a look at these three female genotypes:

Like all females, Females A, B, and C each have two X chromosomes. For Female A, both X chromosomes carry the allele for hemophilia. That means that she is homozygous for the disease and her phenotype will be hemophilia.

Female B is homozygous for the normal allele. She’s not even carrying the allele for hemophilia. Obviously, her phenotype will be normal.

Female C carries the allele for hemophilia on only one of her X chromosomes. Because it is a recessive trait, even though she is carrying the allele, she won’t show it in her phenotype. She won’t have hemophilia. Her phenotype is normal.

What you’ve just learned is this: When it comes to X-linked recessive traits, a female can express the trait in her phenotype only if her genotype is homozygous for the trait. If she’s heterozygous for the trait, then it will not be expressed (it won’t show up) in her phenotype. Females who are heterozygous for X-linked recessive traits are known as carriers. They don’t express the trait, but they do carry the allele for the trait.

Now, what about males? Let’s continue discussing hemophilia, and look at these two male genotypes:

Like all males, Males A and B have an X chromosome (received from their mothers) and a Y chromosome (received from their fathers). Male B is carrying the allele for hemophilia on his X chromosome. Does he have the disease? That is, is his phenotype hemophilia? The answer is yes. Why? Because even though the trait is recessive, there is no normal X chromosome around to dominate it. The male doesn’t have a second X chromosome. He has a Y chromosome instead, and that won’t suppress the expression of the X chromosome’s recessive allele. Male A, of course, is normal. He does not carry the allele for hemophilia.

So, when it comes to an X-linked recessive trait, a female will express the trait in her phenotype only if she is homozygous for it. However, a male will express the trait in his phenotype if he carries the allele on his X chromosome. For this reason, X-linked recessive diseases are much more common in males than in females.

Now, still talking about hemophilia, consider these two parents:

Father Mother

Mother

X(normal)Y

X(hemophilia)X(normal)

Draw a Punnett square and figure out, in terms of sex and hemophilia, what kind of children these two parents might produce. After you’ve done that, fill in the blanks and boxes appropriately.

Genotype Check-In

•   With reference to the parents whose genotypes are shown on this page

(a) _____% of children are likely to be male.

(b) _____% of children are likely to be female.

(c) There is [  no  some ] likelihood that a child will have phenotype hemophilia.

(d) There is [  no  some ] likelihood that a female child will have phenotype hemophilia.

(e) There is [  no  some ] likelihood that a male child will have phenotype hemophilia.

(f) The likelihood that a male child with hemophilia will be born is _____%.

(Answers can be found in Chapter 15.)

Here’s how you figure it out: You draw a Punnett square like this:

The square shows you that the parents are likely to produce:

1.   a male child with genotype XHY phenotype hemophilia

2.   a male child with genotype XY phenotype normal

3.   a female child with genotype XHX phenotype normal

4.   a female child with genotype XX phenotype normal

In our discussion of X-linked traits, we’ve mentioned only ones that are recessive. You’re not likely to be asked questions on the exam about a dominant X-linked trait or a Y-linked trait. Remember, the three X-linked recessive traits you should know about for the SAT Biology E/M Subject Test are hemophilia (which we’ve already discussed), color blindness, and male pattern baldness (color blindness and male pattern baldness are inherited in the same way as hemophilia and follow the exact same rules we discussed for hemophilia). Also, you may not be told that these traits are X-linked; instead, you’ll be expected to know that—and now you do.

One Last Point About X-linked Recessive Traits: They Can’t Go From a Male to a Male

If the test presents you with a question about an X-linked recessive trait, realize this simple truth:


An X-linked recessive trait cannot go from father to son.


That’s because the allele for an X-linked trait is located on the X chromosome, and a father does not give his son an X chromosome. He gives him a Y chromosome.

Pedigree Analysis

The SAT Biology E/M Subject Test might ask you to determine probabilities by reading (analyzing) a pedigree. A pedigree is simply a chart that shows the presence of a particular phenotype in a given family, usually over several generations.

In pedigrees, males are represented by a square, and females are represented by a circle. A horizontal line connecting the two shapes shows the mating between two individuals, and a vertical line connects offspring resulting from that mating. For example, in the pedigree above, individuals 1 and 2 have produced seven offspring: individuals 3, 4, 5, 7, 8, 9, and 10. Individuals 6 and 7 have produced five offspring: individuals 12, 13, 14, 15, and 16. Individuals 10 and 11 have produced four offspring: individuals 17, 18, 19, and 20.

Individuals affected by color blindness are shaded. In our example above, individuals 3, 7, and 18 are affected by the condition. Individuals who marry into the family (in other words, those who are not blood-related) are completely normal and do not carry any abnormal alleles. In our example, these would be individuals 6 and 11.

Questions about pedigrees usually ask you to determine whether a particular individual is heterozygous or homozygous, about the specific genotype of a particular individual, or about the probability of two individuals mating and producing affected offspring. Before attempting to figure out genotypes or probabilities, you must answer two simple questions:

1.   Is the condition dominant or recessive?

Recessive conditions skip generations. Pick any affected individual in the pedigree. Look back at that individual’s parents. Do they have the condition? Look at that individual’s offspring (if present). Do they have the condition? If neither the parents nor the offspring display the condition, the condition is recessive.

Consider our pedigree. Pick individual 7. He is affected by the condition, but neither his parents nor his offspring are affected. This condition is probably recessive. Let’s confirm by looking at another affected individual. Pick individual 18. His parents are not affected and he has no offspring. Because his parents are not affected, this condition is recessive. Of course, if either the parents or the offspring of an affected individual ARE affected, the condition is dominant.

2.   Is the condition sex-linked, or is it autosomal (carried on one of the non-sex chromosomes)?

Sex-linked diseases, as we mentioned earlier, are more common in males than in females. So to determine if the condition is sex-linked, simply count the number of males and females who are affected. If there are significantly more males than females affected, the condition is sex-linked.

Looking back at our pedigree, we see that three males display the condition, and no females are affected. So this condition is sex-linked.

Again, as mentioned earlier, for the purposes of this exam, all sex-linked conditions will be X-linked (carried on the X chromosome). So this condition is not only sex-linked, but, specifically, X-linked.

So then, the condition shown in our pedigree is an X-linked recessive condition. Knowing that, you can answer some basic questions about the condition. (You might want to review the basics on X-linked conditions first.)

1.   Is individual 2 homozygous or heterozygous for the condition?

Individual 2 is female and is not affected by the condition. Therefore, she CANNOT be homozygous, because homozygous females would display a recessive X-linked condition. Furthermore, the fact that she has offspring who are affected by the condition indicates that she must be a carrier for the condition. Therefore, she is heterozygous.

2.   Is individual 13 homozygous or heterozygous for the condition?

Individual 13’s father (individual 7) is affected by this condition. Because fathers pass their X chromosome on to their daughters, and because individual 7’s X chromosome obviously carries the allele for this condition, his daughter, individual 13, must have received the allele for the condition. However, because she is not affected by the condition (she does not display the affected phenotype), she must be heterozygous and a carrier for the condition.

Notice that the same reasoning can be applied to individual 14. All daughters of males affected by an X-linked recessive condition are carriers for the condition.

Following a Trend

The study of the allele
frequency distribution and
change as a result of
migration, natural
selection, genetic drift, or
mutation is called
population genetics.

3.   If individual 13 has a child with a normal male, what is the probability they would produce an affected son?

We’ve already determined that individual 13 is heterozygous for the condition. In other words, her genotype is XaffectedXnormal. A normal male would have the genotype XnormalY. Let’s draw a Punnett square to determine the probability.

The genotype of an affected son would be XaffectedY. From the Punnett square, we see there is only one such genotype out of the four possible genotypes, so the probability of producing an affected son would be 1/4, or 25%.

4.   If individual 3 has a child with a normal female, what is the probability they would produce an affected son?

If you remember the rule about father-to-son transmission of an X-linked recessive condition, this question is easy. The probability is zero (0%), because fathers cannot pass X-linked conditions on to their sons. Fathers give their sons a Y chromosome, not an X. Here’s the Punnett square to prove it:

As you can see, all male children will be normal. (And all female children will be carriers; see the explanation to question 2, above.)

Quick Quiz #3

Consider the pedigree below, then fill in the blanks and check the appropriate boxes:

  1. This condition is [  dominant  recessive ].

  2. This condition is [  autosomal  X-linked ].

  3. The genotype of individual 10 is [  homozygous  heterozygous ].

  4. The genotype of individual 17 is [  homozygous  heterozygous ].

  5. The genotype of individual 1 and 2 are [  homozygous  heterozygous ].

  6. If individual 3 were marry a normal male who does not carry the allele for the condition, the probability they would produce affected offspring would be _______________.

  7. If individual 4 were to marry a female carrier of the condition (a female heterozygous for the condition), the probability they would produce affected offspring would be _______________.

Correct answers can be found in Chapter 15.

Key Words

phenotype

genotype

dominant

recessive

allele

homozygous

heterozygous

codominance

Punnett square

sex chromosomes

autosomes

Mendelian genetics

Gregor Mendel

Mendel’s three laws

  law of dominance

  law of segregation

  law of independent assortment

monohybrid cross

dihybrid cross

test cross

sex-linked trait

X-linked

hemophilia

X-linked recessive trait

carriers

color blindness

male pattern baldness

pedigree

population genetics

Summary

•   An allele is an alternative form, or version, of a gene.

•   Some alleles are dominant; when they exist in a heterozygous individual they appear to be the only one affecting a trait. Some alleles are recessive. When they exist in a heterozygous individual, the trait that they code for is suppressed.

•   Crosses (such as those represented by Punnett squares) can be used to predict the phenotype and genotype of the offspring of two particular parents.

•   One of the 23 pairs of chromosomes in humans determines the gender of the offspring. These are called sex chromosomes. The rest of the chromosomes are called autosomes.

•   Gregor Mendel is the father of genetics and his studies in pea plants and true-breeding resulted in his theories which are known as Mendelian genetics.

•   Some alleles are passed from generation to generation only on the sex chromosomes.

•   Pedigrees are diagrams that show the presence of phenotypes in several generations of a family.