Genetics Practice Problems

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Monohybrid Cross:

In humans, brown eyes (B) are dominant over blue (b)*. A brown-eyed man marries a blue-eyed woman and they have three children, two of whom are brown-eyed and one of whom is blue-eyed. Draw the Punnett square that illustrates this marriage. What is the man’s genotype? What are the genotypes of the children?

(* Actually, the situation is complicated by the fact that there is more than one gene involved in eye color, but for this example, we’ll consider only this one gene.)

Testcross:

In dogs, there is an hereditary deafness caused by a recessive gene, “d.” A kennel owner has a male dog that she wants to use for breeding purposes if possible. The dog can hear, so the owner knows his genotype is either DD or Dd. If the dog’s genotype is Dd, the owner does not wish to use him for breeding so that the deafness gene will not be passed on. This can be tested by breeding the dog to a deaf female (dd). Draw the Punnett squares to illustrate these two possible crosses. In each case, what percentage/how many of the offspring would be expected to be hearing? deaf? How could you tell the genotype of this male dog? Also, using Punnett square(s), show how two hearing dogs could produce deaf offspring.

Incomplete Dominance:

Note: at least one textbook I’ve seen also uses this as an example of pleiotropy (one gene – multiple effects), though to my mind, the malaria part of this is not a direct “effect” of the gene.

For many genes, such as the two mentioned above, the dominant allele codes for the presence of some characteristic (like, “B” codes for “make brown pigment” in someone’s eyes), and the recessive allele codes for something along the lines of, “I don’t know how to make that,” (like “b” codes for the absence of brown pigment in someone’s eyes, so by “default,” the eyes turn out blue). If someone is a heterozygote (Bb), that person has one set of instructions for “make brown” and one set of instructions for, “I don’t know how to make brown,” with the result that the person ends up with brown eyes. There are, however, some genes where both alleles code for “something.” One classic example is that in many flowering plants such as roses, snapdragons, and hibiscus, there is a gene for flower color with two alleles: red and white. However, in that case, white is not merely the absence of red, but that allele actually codes for, “make white pigment.” Thus the flowers on a plant that is heterozygous have two sets of instructions: “make red,” and “make white,” with the result that the flowers turn out mid-way in between; they’re pink.

In humans, there is a gene that controls formation of hemoglobin, the protein in the red blood cells which carries oxygen to the body tissue. The “normal” allele of this gene codes for “normal” hemoglobin. However, there is another allele for this gene that has one different nitrogenous base in its DNA sequence, and thus, one codon in the middle of the gene codes for a different amino acid in an important place in the hemoglobin molecule. A red blood cell (RBC) that contains this altered hemoglobin will, under stress, crinkle up into a shape that reminded someone of the shape of an old-fashioned sickle. While the letters “S” and“s” are often used to represent these alleles, since both of them code for “make hemoglobin”, in reality, neither is dominant over the other. Someone who is SS makes all normal hemoglobin, someone who is ss makes all abnormal hemoglobin (and we say that person has sickle-cell anemia), and someone who is Ss essentially has two sets of instructions, and so, makes some of each kind of hemoglobin (often referred to as sickle-cell trait).

Because the RBCs of a person who is ss contain all abnormal hemoglobin, they will “sickle” very easily, with very little stress required to provoke that reaction. All those sickled cells tend to get stuck as they try to go through capillaries, and cause things like strokes, heart attacks, pulmonary embolisms, etc. that lead to death. Because only some of the RBCs of a person who is Ss contain abnormal hemoglobin, that person usually only has trouble with a lot of cells sickling if they’re under a lot of stress trying to meet a higher-than-normal oxygen demand, and so the chances of a person dying from sickle-cell trait are much lower than for full-blown sickle-cell anemia.

A photo, taken by Dr. Fankhauser, of a prepared slide of blood cells infected with Plasmodium vivax

Malaria is a parasitic disease that’s prevalent in tropical areas. When a mosquito that’s carrying the parasites bites someone, the parasites enter the person’s bloodstream, and invades and lives in the person’s RBCs. However, if a person has sickle-cell anemia (ss), the presence of a parasite in a RBC is so stressful, it causes the RBC to sickle (crinkle up), and when that happens, that kills the parasite before it can multiply and spread to other RBCs. Thus, coincidentally, a person who is ss is also “immune” to malaria. If a person is Ss and a malaria parasite tries to invade a RBC with abnormal hemoglobin, again, the RBC will sickle, killing the parasite before it has a chance to reproduce. If a parasite invades a RBC with normal hemoglobin, it will be able to live and multiply, but if its offspring invade other RBCs with abnormal hemoglobin, they, too, will be killed. Thus, a person who is Ss is “resistant” (though not totally immune) to malaria. If a person is SS and has all normal hemoglobin, the malaria parasites do just fine, invading RBCs, growing and multiplying, and invading more RBCs. Thus, an SS person usually dies, eventually, from causes tied to the malaria.

A man and woman living in a tropical area where malaria is prevalent and health care is not accessible have seven children. The genotypes of these children are ss, Ss, SS, ss, Ss, Ss, and SS.

Dihybrid Cross:

In humans, there is a gene that controls formation (or lack thereof) of muscles in the tongue that allow people with those muscles to roll their tongues, while people who lack those muscles cannot roll their tongues. The ability to roll one’s tongue is dominant over non-rolling. The ability to taste certain substances is also genetically controlled. For example, there is a substance called phenylthiocarbamate (PTC for short), which some people can taste (the dominant trait), while others cannot (the recessive trait). The biological supply companies actually sell a special kind of tissue paper impregnated with PTC so students studying genetics can try tasting it to see if they are tasters or non-tasters. To people who are tasters, the paper tastes very bitter, but to non-tasters, it just tastes like paper. Let’s let R represent tongue-rolling, r represent a non-roller, T represent ability to taste PTC, and t represent non-tasting.

Suppose a woman who is both a homozygous tongue-roller and a non-PTC-taster marries a man who is a heterozygous tongue-roller and is a PTC taster, and they have three children: a homozygous tongue-roller who is also a PTC taster, a heterozygous tongue-roller who is also a taster, and a heterozygous tongue-roller who is a non-taster. If these parents would have a bunch more children so that they had 12 in all, how many of those 12 would you expect to be non-tasters who are homozygous for tongue-rolling? If the first child (the homozygous tongue-roller who is also a PTC taster) marries someone who is heterozygous for both traits, draw the Punnett square that predicts what their children will be.

Multiple Alleles and Codominance:

Some genes have more than two alleles. One of the best-known examples is the gene that is referred to as the “ABO Blood Group,” which actually has quite a number of alleles. However, we will discuss/consider only the three most-common of these. This gene codes for the structure of a certain antigen on the surface of our RBCs. The three alleles we will work with are symbolized by I^A, I^B, and i. However, keep in mind that a person can only have two alleles, two copies of a gene. Thus, the possible genotypes are I^AI^A, I^Ai, I^BI^B, I^Bi, I^AI^B, or ii. (Sometimes, you will see these simplified as AA, AO, BB, BO, AB, and OO, but that does make it harder to remember that these are all alleles for the same gene.)

The allele, I^A, codes for, “make type A antigen,” the allele I^B codes for, “make type B antigen,” and (to simplify things somewhat) the i allele codes for, “I don’t know how to make either A or B.” Thus, both I^AI^A and I^Ai individuals receive instructions to “make type A antigen,” and both I^BI^B and I^Bi individuals receive instructions to “make type B antigen.” Individuals who are I^AI^B receive two sets of instructions: “make type A” and “make type B,” so they have both the A and B forms of that antigen on the surface of their RBCs. People who are ii don’t have any instructions to make either A or B, so by “default” they make what we refer to as type O antigens. Since both I^A and I^B code for “make something” whereas i codes for, “I don’t know how, therefore, both I^A and I^B are dominant over i. However, since both I^A and I^A code for “make something,” neither of them is dominant over the other. Thus we say that I^A and I^B are codominant over i.

Suppose a person with type A blood and a person with type B blood get married. What are the possible genotypes their children could have?

There is another gene that codes for another, different antigen that also occurs on the surface of our RBCs, and technically, that gene also has multiple alleles. However, most people either have or do not have one particular allele called the “d” allele. This gene codes for an antigen that is called “Rh factor” because it was first discovered in Rhesus monkeys. People who have instructions to “make d antigen” are referred to as Rh⁺ (the allele is often symbolized by the letter “R”), while those who have “I don’t know how to make d antigen” instructions are called Rh^– (the allele can be symbolized by “r”). Since this is a totally separate gene than the ABO blood group, if you’re doing a genetic cross that involves both ABO and Rh, that would be a dihybrid cross.

Trihybrid Cross:

In Guinea pigs, black hair (B) is dominant over white (b), rough coat texture (R) is dominant over smooth (r), and short hair (S) is dominant over long hair (s). Assuming these genes are on separate chromosomes, draw the Punnett square for a cross between a homozygous black, rough, short-haired Guinea pig and a white, smooth, long-haired one. What would the phenotype(s) of the offspring be? If two of the F₁ offspring were crossed, draw the Punnett square for this cross. Hint: first make a list of the possible gametes, making sure each has exactly one copy of each of the genes (one allele for each gene). What would the genotype and phenotype ratios be for the F₂ generation?

Polygenic Trait:

Some traits, some phenotypes, are controlled by more than one gene. It was mentioned in the monohybrid cross, above, that technically, human eye color is controlled by at least two genes, one which codes for brown vs. blue and another which codes for green vs. blue. In the epistasis crosses, below, you will see other examples of polygenic traits. Human skin color is also a classic example of a polygenic trait. It is known that at least three or four genes control skin color, and for each of those genes, dark pigment has incomplete dominance over light (so a heterozygote would be intermediate — see above). Because we just did a trihybrid cross, let’s assume three genes here (for simplicity), and to avoid confusion among them, let’s arbitrarily call them genes A, B, and C. Then, someone who is AABBCC would have very dark skin color and someone who is aabbcc would have very light skin color. If they would get married and have children, their children would all be AaBbCc. If two of those people would get married and have children, the Punnett square would look like this:

Linked Genes:

Normal fruit flies have grayish-yellow bodies, red eyes, and wings that are long-enough to be able to fly. Some mutant fruit flies have black bodies rather than grayish-yellow, some have stumpy, vestigial wings that are too short and wrinkled to be able to fly, and some have a brighter, orangish-red eye color that is called “cinnabar.” By breeding flies, fruit fly researchers were able to determine that all three of these mutations were recessive and were on the autosomes.

Fruit fly researchers use a different type of symbolism to represent their genetic crosses. They use a plus sign (+) to indicate anything that is the “wild” type and a letter (or sometimes two) to represent a mutant allele (capital if the mutation is a dominant allele, lower case if it’s recessive). Thus a fruit fly with a homozygous grayish-yellow body would be labeled as “++” for body color while a black-bodied fly would be “bb” and a heterozygous fly would be “+b”. The genotype for normal wings would also be symbolized as “++,” while the allele for vestigial wings is “vg” so a fly that is homozygous recessive for that gene would be “vgvg” and a heterozygote would be notated as “+vg”. Similarly, normal red eyes would be “++,” cinnabar eyes would be ”cncn,” and a heterozygote would be “+cn.”

wild-type female		wild-type male
black, vestigial female		black, vestigial male

To simplify things, for now, let’s consider just the b and vg genes. Suppose we had some flies that were genotype +b+vg (grayish body and long wings, but a carrier for both the black and vestigal alleles). Let’s do a testcross (= a cross with bbvgvg, remember?) with these heterozygous flies.

If these genes were on different chromosomes, the Punnett square for this cross would look like this:

	++	+vg	b+	bvg
bvg	+b+vg +bvgvg bb+vg bbvgvg

so out of the offspring, we would expect to get ¼ of each of the four types.

If the genes were linked on the same chromosome and one of the fly’s chromosomes contained the alleles for yellowish body and normal wings while the other chromosome contained the alleles for black body and crumpled, vestigial wings, the Punnett square for this cross would look like this:

	++	bvg
bvg	+b+vg bbvgvg

so out of the offspring, we would expect to get ½ of each of the two types.

Sex-Linked Genes:

In humans, the genes for colorblindness and hemophilia are both located on the X chromosome with no corresponding gene on the Y. These are both recessive alleles. If a man and a woman, both with normal vision, marry and have a colorblind son, draw the Punnett square that illustrates this. If the man dies and the woman remarries to a colorblind man, draw a Punnett square showing the type(s) of children could be expected from her second marriage. How many/what percentage of each could be expected?

Sex (It’s a Phenotype, Not a Chromosome!):

In humans, there’s another X-linked gene that codes for androgen (testosterone) receptors in our cells. The dominant allele (X^A) codes for “make functioning receptors that can correctly receive and bind onto testosterone,” but there is a recessive allele (X^a) that either codes for “I don’t know how to make testosterone receptors” or else “make ‘broken’ receptors that can’t receive and bind onto testosterone.” In humans, there is also a Y-linked gene that codes for “make testes,” and when present, they, in turn, make testosterone, and the testosterone, in turn, goes to other cells in the body, and when received by the receptors, triggers other events within those cells. During embryonic development, one such event is growth and development of male genitalia.

Consider a person who is genotype X^AY. Because this person has a Y chromosome including a normally-functioning gene to make testes, at the appropriate time during embryonic development, testes will form and will start to secrete testosterone. Because this person also has the correctly-functioning allele for the androgen (testosterone) receptor gene, those receptors will form and will begin functioning. As they receive the testosterone made by the testes, this will stimulate development of male genitalia, and (assuming all other genes are working “normally”), this baby will be a boy.

Consider a person who is genotype X^AX^A. Because this person does not have a Y chromosome, there is no gene to provide instructions to make testes, therefore no big prenatal surge of testosterone, therefore no stimulus to make male genitalia (in spite of properly-functioning testosterone receptors), so by “default,” female genitalia develop, and (assuming all other genes are working “normally”), this baby will be a girl.

If these two people would get married, the Punnett square for their children would look pretty much like the ones you’ve just done:

	XA	Y
XA	XAXA XAY XAXA XAY
XA

Consider a person who is genotype X^AX^a. Because this person does not have a Y chromosome, there is no gene to provide instructions to make testes, therefore no big prenatal surge of testosterone, therefore no stimulus to make male genitalia (in spite of properly-functioning testosterone receptors in most cells), so by “default,” female genitalia develop, and (assuming all other genes are working “normally”), this baby will also be a girl. However, she is a carrier for the non-functioning androgen-receptor gene. Because of inactivation of X chromosomes to form Barr bodies, some of the cells in her body will not be able to receive testosterone messages, which might show up in things like having somewhat less pubic and/or armpit hair.

Sex-Influenced Traits:

Baldness in humans is a dominant, sex-influenced trait. This gene is on the autosomes, not the sex chromosomes, but how it is expressed is influenced by the person’s sex (due to hormones present, etc.). A man who is BB or Bb will be bald and will be non-bald only if he is bb. A woman will only be bald if she is BB and non-bald if she is Bb or bb (it’s almost like B is dominant in males and b is dominant in females). Actually, because of the influence of other sex-related factors, most women who are BB never become totally bald like men do, but rather, their hair becomes “thin” or sparse. If two parents are heterozygous for baldness, what are the chances of their children being bald? Use a Punnett square to illustrate this. Note: because the sex of a person does make a difference in how the gene is expressed, you need to set this up as a dihybrid cross to account for the sex of the children.

A non-bald man marries a non-bald woman. They have a son and a daughter. If the son becomes bald, what are the chances that his sister will, too? Use a Punnett square to show this cross.

A woman’s mother is bald, but her father is not. Her older brother is rapidly going bald. She is an acrobat who hangs by her hair. Should she change her profession before she goes bald, too? Use a Punnett square to show this.

Epistasis:

	In sweet peas, purple flower color (P) is dominant over white (p), but there is also a control gene such that if the plant has a “C”, the purple has “permission” to express itself. If the plant is “cc,” the purple does not “have permission” to express itself and the flower will be white anyway. If a plant with homozygous purple, controlled flowers is crossed with a plant with white, non-controlled flowers, diagram the Punnett square for the F1 and F2 generations and calculate the genotype and phenotype ratios.

Epistasis can also work in the “opposite” direction (actually, some textbooks give these two cases other names). In corn kernels, purple (P) is dominant over yellow (p), but there is also an inhibiting control gene such that if that corn kernel has a “C”, the purple does not “have permission” to express itself, and the kernel will be yellow. If that kernel is “cc”, then the purple has “permission” to be expressed. If the same cross as above is done, the result would be:

genotypes:					phenotypes:
PPCC	1				P–C–	= yellow bec. of C–	9	}	13 yellow total
PPCc	2				ppC–	= yellow	3
PPcc	1				ppcc	= yellow bec. of pp	1
PpCC	2				P–cc	= purple	3	}	3 purple
PpCc	4
Ppcc	2
ppCC	1
ppCc	2
ppcc	1

There are some other, interesting possibilities. Suppose that in corn kernels, purple (P) is dominant over some other color, say red (p). However, suppose there is a control gene (C) which “permits” the color to be present and “cc” doesn’t “permit” color to be present, resulting in yellow kernels. If the same cross as above is done, the result would be:

genotypes:					phenotypes:
PPCC	1				P–C–	= purple	9	}	9 purple
PPCc	2				P–cc	= yellow bec. of cc	3	}	4 yellow total
PPcc	1				ppcc	= yellow bec. of cc	1
PpCC	2				ppC–	= red bec. of pp	3	}	3 red
PpCc	4
Ppcc	2
ppCC	1
ppCc	2
ppcc	1

Suppose, instead, that “C” inhibits or does not “permit” the color to be present, while “cc” does allow the color to be expressed. If the same cross as above is done, the result would be:

genotypes:					phenotypes:
PPCC	1				P–C–	= yellow bec. of C–	9	}	12 yellow total
PPCc	2				ppC–	= yellow bec. of C–	3
PPcc	1				P–cc	= purple	3	}	3 purple
PpCC	2				ppcc	= red bec. of pp	1	}	1 red
PpCc	4
Ppcc	2
ppCC	1
ppCc	2
ppcc	1

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Copyright © 1996 by J. Stein Carter. All rights reserved.
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