Genetics Practice Problems
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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.)
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.
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.
In humans, there is a gene that controls formation (or lack
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.
If the man is both Rr and Tt (How do we know that?), he would be RrTt and so could produce gametes with either R or r and either T or t (one allele for each gene). There are two choices for the first trait (R or r). No matter which of those go into a given sperm, there are still two choices for the second trait (T or t). Therefore, a total of 2 × 2 = 4 possible types of gametes (sperm) may be produced.
Multiple Alleles and Codominance:
Some genes have more than two alleles. One of the best-known
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 IA, IB, and i. However, keep in mind that a person
can only have two alleles, two copies of a gene. Thus, the possible genotypes
are IAIA, IAi, IBIB,
IBi, IAIB, 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, IA, codes for, “make type A antigen,” the allele IB 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 IAIA and IAi individuals receive instructions to “make type A antigen,” and both IBIB and IBi individuals receive instructions to “make type B antigen.” Individuals who are IAIB 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 IA and IB code for “make something” whereas i codes for, “I don’t know how, therefore, both IA and IB are dominant over i. However, since both IA and IA code for “make something,” neither of them is dominant over the other. Thus we say that IA and IB 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.
Ms. Johnston, Ms. Johnson, and Ms. Johnstone all entered the same hospital and gave birth to baby girls on the same day, and all three babies were taken to the nursery to receive care, there. Someone later claimed that the hospital mixed up the babies. As a hospital administrator, it is your job to make sure that each pair of parents has the correct baby, so you order blood typing to be done on all the parents and all the babies. Here are the results:
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?
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 AaBbCc people would get married and have children, the Punnett square would look like this:
|black, vestigial female|
|black, vestigial male|
If these genes were on different chromosomes, the Punnett square for this cross would look like this:
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:
so out of the offspring, we would expect to get ½ of each of the two types.
In humans, the genes for colorblindness and hemophilia are both located on the X chromosome with no corresponding gene on the Y. Typically an X chromosome with a “normal” allele is notated as a plain X while an X chromosome carrying the mutant allele is notated as X′ or with an appropriate letter, such as Xc for colorblindness or Xh for hemophilia (thus, while it’s not typically used, you could use XC or XH to represent the normal allele, especially if you’re dealing with both, simultaneously). Note that besides being sex-linked genes, hemophilia and colorblindness are also linked genes and a man who has both would be XchY. 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
receptors in our cells. The dominant allele (XA) codes for “make
functioning receptors that can correctly receive and bind onto testosterone,”
but there is a recessive allele (Xa) 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
Consider a person who is genotype XAY. 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 XAXA. 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:
½ of the children will be females, and ½ will be males
Consider a person who is genotype XAXa. 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 many/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.
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 may become “thin” or sparse.
If two parents are heterozygous for baldness (note: that means he would be
bald, but she would not.), 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.
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
Because you must keep track of both the purple/white and the control genes, you need to set this up as a dihybrid cross.
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:
|PPCC||1||P–C–||= yellow bec. of C–||9||}||13 yellow total|
|PPcc||1||ppcc||= yellow bec. of pp||1|
|PpCC||2||P–cc||= purple||3||}||3 purple|
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:
|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|
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:
|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|
ZW Sex-Linked Inheritance:
Birds and lepidopterans (butterflies and moths) have ZW sex
determination, which is so named to distinguish it from XY sex determination
(as in humans and fruit flies) because ZW is the “opposite” of XY. In the ZW
sex determination system, the females are heterogametic (ZW), and the males
are homogametic (ZZ).
Barred Rooster In chickens, there is an autosomal gene called “extended black,” symbolized by E. There is also a sex-linked (Z-linked) dominant gene for barred feathers (causing the feathers of the adults to be black and white striped/barred). Since this is a sex-linked gene, the barred allele would be symbolized as ZB, and the non-barred allele as Zb.
A genetic cross that is routinely done in the poultry industry is to cross a Rhode Island Red or New Hampshire breed rooster, both of which are EEZbZb for these two genes with a Barred Plymouth Rock hen, which breed is EEZBW for these two genes. This results in hybrid offspring of a variety (not a true-breeding “breed”) called Black Sex-Linked. When the new Black Sex-Linked chicks hatch, they can be sexed immediately because the females are all black while the males have a cream-colored or whitish spot on their heads. As adults, the hens will be solid-colored while the roosters will have barred feathers.
For purposes of this illustration, since both parents are EE, we can “ignore” that gene, and look just at the sex-linked (Z-linked) barring gene. Thus, the cross to produce Black Sex-Linked chicks can be represented by ZBW × ZbZb.
The resulting offspring would be ZbW hens (called pullets when young) and ZBZb roosters (called cockerels when young). It is the ZB allele inherited by the roosters that gives them their white head spots as chicks and also give them barred feathers as adults.
If one each of those Black Sex-Linked hens and roosters would manage to live to adulthood and mate with each other:
Thus, with the probability of ¼ of each of four possible types of offspring, clearly this hybrid variety of chickens is not true-breeding.
Copyright © 1996, 2011 by J. Stein Carter. All rights reserved.