Genetics Practice Problems

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Monohybrid Cross:

Family 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.)

If blue is recessive, what must the woman’s genotype be?


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If that’s her genotype, what kind(s) of gametes (eggs) can she produce?

and/or
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If the man has brown eyes, but has a blue-eyed child what must his genotype be?
(if you don’t understand why, review the testcross problem)


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If that’s his genotype, what kind(s) of gametes (sperm) can he produce?

and/or
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If they have children:

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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½ of their children would be expected to have heterozygous brown eyes, and ½ homozygous blue.


Testcross:

Dog 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.

A deaf female is genotype dd. What kind(s) of gametes (eggs) can she produce?

and/or
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If the hearing male is DD, what kind(s) of gametes (sperm) can he produce?

and/or
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If he is Dd, what kind(s) of gametes can he produce?

and/or
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If he is DD:

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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All (100%) of the offspring can hear.

If he is Dd:

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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½ of the offspring can hear, and ½ of the offspring are deaf.

If any of the puppies are deaf, the male is Dd, but assuming a litter of at least about 4 to 5 puppies or more, if they can all hear, he is most likely DD.

If two hearing dogs were both Dd, what kind(s) of gametes (eggs/sperm) could each produce?

and/or
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If they are bred with each other:

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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We would expect ¾ of the puppies to be normal and ¼ deaf.


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.

Plasmodium vivax in RBCs
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.

What must the genotype of both parents be?
(Hint: what would be needed to have those kinds of children?)


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If that’s their genotype, what kind(s) of gametes (eggs/sperm) can each produce?

and/or
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To figure out what their children would be:

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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¼ of their children would be expected to have normal hemoglobin, ½ should have sickle-cell trait, and ¼ should have full-blown sickle-cell anemia.

Which of their children would you expect to live to adulthood and reproduce?

ssSsSSssSsSsSS
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What is the most likely genotype of the children’s future spouses?


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What is the most likely genotype of the surviving grandchildren?


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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.

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.

What kind(s) of gametes (sperm) can he produce?

and/or and/or and/or
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The woman would be RRtt (How do we know that?).

What kind(s) of gametes (eggs) can she produce?

and/or and/or and/or
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We could put four rows of female gametes, but to simplify things, since they’re all the same, we can get by with one row.

*   * *

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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The three children marked with an asterisk (*) are the children mentioned in the problem. ¼ of the children (3 out of 12) would be expected to be RRtt (homozygous tongue-rollers who are non-tasters).

What kind(s) of gametes (let’s say this is a daughter so the gametes are eggs) can their first child (the homozygous tongue-roller who is also a PTC taster) produce?

and/or and/or and/or
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Her husband, who is heterozygous for both traits, would be RrTt like her father. What kind(s) of sperm can her husband produce?

and/or and/or and/or
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If they have children:

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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Notice that the children will be ½ homozygous tongue-rollers, ½ heterozygous tongue-rollers;
¾ of the children will be PTC tasters and ¼ non-tasters;
¾ of the homozygous tongue-rollers will be PTC tasters and ¼ non-tasters;
¾ of the heterozygous tongue-rollers will be PTC tasters and ¼ non-tasters;
⅜ of the children will be homozygous tongue-rollers who are PTC tasters,
⅜ heterozygous tongue-rollers who are PTC tasters,
⅛ homozygous tongue-rollers who are non-tasters, and
⅛ heterozygous tongue-rollers who are non-tasters.


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 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?

First, what possible genotypes can a person be if that person has type A blood?
(Hint: to keep the computer happy, use A, B, and O, rather than IA, IB, and i, even though the latter are more correct.)

and/or
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Which of those genotypes would give rise to the wider variety of genotypes in the children that are produced?


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If that’s that person’s genotype, what kind(s) of gametes (eggs/sperm) would that person produce?

and/or
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What possible genotype(s) can a person be if that person has type B blood?

and/or
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Which of those genotypes would give rise to the wider variety of genotypes in the children that are produced?


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If that’s that person’s genotype, what kind(s) of gametes (eggs/sperm) would that person produce?

and/or
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If the mother is the type A person and the father is the type B person,

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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¼ of their children would have type A blood, ¼ would have type B blood,
¼ would have type AB blood,
and ¼ would have type O blood.

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:

Person Blood Type
(Phenotype)
Probable
Genotype
Possible Gametes
(Eggs/Sperm)
Ms. Johnston A+
Mr. Johnston B+
Ms. Johnson B
Mr. Johnson O+
Ms. Johnstone A+
Mr. Johnstone A
Baby A O+  
Baby B AB
Baby C B

First, for each parent, think about what possible genotype(s) could give that phenotype. If there’s more than one possible genotype, then which of the possible genotypes would give the most variation in terms of possible children?

Put that genotype in the appropriate box, above, for each person.
(Hint #1: remember, this is a dihybrid cross)
(Hint #2: to keep the computer happy, use letters like A, B, O, R, and/or r)

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Next, for each parent, if that is his/her genotype, then what possible kinds of gametes can each produce? Put those in the appropriate boxes, above.
(Hint: not everyone makes four different types of gametes, so there are some extra boxes that you should leave blank.)

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Then, for each pair of parents, do a Punnett square to determine what kinds of children they can have.

Johnston:

Johnson:

Johnstone:

For each of the three Punnett squares, put the female gametes (eggs) in the correct locations.

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For each of the three Punnett squares, put the male gametes (sperm) in the correct locations.

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For each of the three Punnett squares, put the genotypes of the offspring in the correct locations.

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So, now, who are the parents of which baby?

 Baby A 
 O+ 
 Baby B 
 AB 
 Baby C 
 B 
 Johnston 

 Johnson 

 Johnstone 


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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?

In the parental generation,
if we arbitrarily let the female parent be BBRRSS, what kind(s) of gametes (eggs) can she produce?

and/or and/or and/or and/or
and/or and/or and/or
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If we arbitrarily let the male parent be bbrrss, what kind(s) of gametes (sperm) can he produce?

and/or and/or and/or and/or
and/or and/or and/or
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Thus, all of the F1 will have the same genotype. What will that be?


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What will they look like – what will their phenotype be?
(Hint: to keep the computer happy, use a comma and space to separate multiple words.)


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What kind(s) of gametes (eggs/sperm) can they produce?

and/or and/or and/or and/or
and/or and/or and/or
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If they are bred with each other:

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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genotypes: phenotypes:
BBRRSS 1 BbRRSS 2 bbRRSS 1 B–R–S– 27
BBRRSs 2 BbRRSs 4 bbRRSs 2 B–R–ss 9
BBRRss 1 BbRRss 2 bbRRss 1 B–rrS– 9
BBRrSS 2 BbRrSS 4 bbRrSS 2 B–rrss 3
BBRrSs 4 BbRrSs 8 bbRrSs 4 bbR–S– 9
BBRrss 2 BbRrss 4 bbRrss 2 bbR–ss 3
BBrrSS 1 BbrrSS 2 bbrrSS 1 bbrrS– 3
BBrrSs 2 BbrrSs 4 bbrrSs 2 bbrrss 1
BBrrss 1 Bbrrss 2 bbrrss 1

To double check, number of squares = 64,
sum of genotypes = 64,
and sum of phenotypes also = 64.


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  AaBbCc  people would get married and have children, the Punnett square would look like this:

ABC

ABc

AbC

Abc

aBC

aBc

abC

abc

ABC

AABBCC

AABBCc

AABbCC

AABbCc

AaBBCC

AaBBCc

AaBbCC

AaBbCc

ABc

AABBCc

AABBcc

AABbCc

AABbcc

AaBBCc

AaBBcc

AaBbCc

AaBbcc

AbC

AABbCC

AABbCc

AAbbCC

AAbbCc

AaBbCC

AaBbCc

AabbCC

AabbCc

Abc

AABbCc

AABbcc

AAbbCc

AAbbcc

AaBbCc

AaBbcc

AabbCc

Aabbcc

aBC

AaBBCC

AaBBCc

AaBbCC

AaBbCc

aaBBCC

aaBBCc

aaBbCC

aaBbCc

aBc

AaBBCc

AaBBcc

AaBbCc

AaBbcc

aaBBCc

aaBBcc

aaBbCc

aaBbcc

abC

AaBbCC

AaBbCc

AabbCC

AabbCc

aaBbCC

aaBbCc

aabbCC

aabbCc

abc

AaBbCc

AaBbcc

AabbCc

Aabbcc

aaBbCc

aaBbcc

aabbCc

aabbcc

So, to summarize, as above:

genotypes: phenotypes:
AABBCC (6 dark) 1 AaBBCC (5 dark) 2 aaBBCC (4 dark) 1 6 dark alleles
AABBCc (5 dark) 2 AaBBCc (4 dark) 4 aaBBCc (3 dark) 2 5 dark alleles
AABBcc (4 dark) 1 AaBBcc (3 dark) 2 aaBBcc (2 dark) 1 4 dark alleles
AABbCC (5 dark) 2 AaBbCC (4 dark) 4 aaBbCC (3 dark) 2 3 dark alleles
AABbCc (4 dark) 4 AaBbCc (3 dark) 8 aaBbCc (2 dark) 4 2 dark alleles
AABbcc (3 dark) 2 AaBbcc (2 dark) 4 aaBbcc (1 dark) 2 1 dark allele
AAbbCC (4 dark) 1 AabbCC (3 dark) 2 aabbCC (2 dark) 1 0 dark alleles
AAbbCc (3 dark) 2 AabbCc (2 dark) 4 aabbCc (1 dark) 2                     ↑
How many of
each phenotype?
AAbbcc (2 dark) 1 Aabbcc (1 dark) 2 aabbcc (0 dark) 1

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Linked Genes:

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

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-yellow body and long wings, but carriers 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.

Suppose you are a fruit fly researcher who has bred a group of +b+vg females with bbvgvg males, and now, you have to count all their offspring. How many offspring would you like to count?


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Females: Males:

Based on your results, what can you conclude? Would you say these genes are linked or not? What evidence can you cite to support your conclusion? Actually, in real life, the results are not as simple as this due to something called “crossing-over.”

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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. 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?

A man with normal vision is XY. What kind(s) of gametes (sperm) can he produce?

and/or
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Any woman with normal vision could be XX or XX′.
Since this woman has a colorblind son (genotype X′Y), she has to be XX′ (a carrier).

What kind(s) of gametes (eggs) can she produce?

and/or
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Her first marriage would look like:

   
     
    *

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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The child marked with an asterisk (*) is the son in the problem.

In order to be colorblind, her second husband must be X′Y (like her son).
What kind(s) of gametes (sperm) can he produce?

and/or
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Her second marriage would look like:

Put the female gametes in the correct locations.

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¼ of the children will be normal female carriers, ¼ colorblind females,
¼ normal males, and ¼ colorblind males;
½ of the children will be females and ½ males;
½ of the children will have normal vision and ½ will be colorblind;
½ of the daughters will be carriers and ½ colorblind;
½ of the sons will be normal and ½ colorblind.


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 (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 male genitalia.

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:

XA

Y

XA

XAXA

XAY

XA

XAXA

XAY

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.

If she marries an XAY man like the one mentioned above, what would their children’s genotypes be?

XA

Y

XA

Xa

Put the genotypes of the offspring in the correct locations.

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¾ of the children will be females, and ¼ will be males

No, that’s not a typo! The phenotype ratio is 75:25, not 50:50. There’s a “new” genotype in there that hasn’t been mentioned, yet: XaY, and the corresponding phenotype is female. Let’s look at that person’s embryonic development. There is a Y chromosome, and assuming its genes are functioning normally, testes do develop and begin to secrete testosterone. However, all that testosterone has nowhere to go; the testosterone receptors don’t work, so as far as everything else in her body is concerned, its as though all that testosterone isn’t even there. Thus, just exactly like anyone/everyone else whose body doesn’t get a message/stimulus to make male genitalia, by “default” female genitalia develop, and (assuming all other genes are working “normally”), this baby will be a girl. This is called Androgen Insensitivity Syndrome (AIS). See the Sex-Linked Genes Web page for a more detailed explanation.


Sex-Influenced Traits:

Bald Man 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.

The woman is XXBb. What kind(s) of gametes (eggs) can she produce?

and/or and/or and/or
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Her husband would be XYBb. What kind(s) of gametes (sperm) can he produce?

and/or and/or and/or
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If they have children:

       
  *   * *
      *  

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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Those marked with an asterisk (*) will be bald.

⅛ of the children will be bald daughters, ⅜ will be normal daughters,
⅜ will be bald sons, and ⅛ will be normal sons;

½ of the children will be daughters and ½ sons;
½ of the children will be bald and ½ normal;
¼ of the daughters will be bald and ¾ normal;
¾ of the sons will be bald and ¼ normal.

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 non-bald man has to be XYbb. What kind(s) of gametes (sperm) can he produce?

and/or and/or and/or
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Any non-bald woman can be XXBb or XXbb. The bald son could be XYBB or XYBb, but since the father is XYbb, we know the son cannot be XYBB (remember the first problem?). The son has to be XYBb, therefore this mother has to be XXBb (if she was XXbb he couldn't have a B). What kind(s) of gametes (eggs) can she produce?

and/or and/or and/or
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If they have children:

   
    *
     

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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The child marked with an asterisk (*) is the son in the problem.
The daughter could be XXBb or XXbb, in neither case, bald: 0% chance.

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.

If the mother is bald, she has to be XXBB. What kind(s) of gametes (eggs) can she produce?

and/or and/or and/or
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If the father is not bald, he has to be XYbb. What kind(s) of gametes (sperm) can he produce?

and/or and/or and/or
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What will the genotype of all their male children (like her brother) be?


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What will the genotype of all their female children (like her) be?


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Will she go bald?

Yes, get a different job. || No, keep hanging by her hair.
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Epistasis:

Purple Pea Flower White Pea Flower 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.

Because you must keep track of both the purple/white and the control genes, you need to set this up as a dihybrid cross.

If we arbitrarily let the female parent be PPCC, what kind(s) of gametes (eggs) can she produce?

and/or and/or and/or
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If we arbitrarily let the male parent be ppcc, what kind(s) of gametes (sperm) can he produce?

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Thus, all of the F1 will have the same genotype. What will that be?


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What will they look like – what will their phenotype be?


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If all of the F1 are PpCc, what kind(s) of gametes (eggs/sperm) can they produce?

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The Punnett square for the F2 generation will look like:

Put the female gametes in the correct locations.

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Put the male gametes in the correct locations.

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Put the genotypes of the offspring in the correct locations.

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genotypes: phenotypes:
PPCC 1 P–C– = purple 9 } 9 purple
PPCc 2 P–cc = white bec. of cc 3 } 7 white total
PPcc 1 ppC– = white bec. of pp 3
PpCC 2 ppcc = white 1
PpCc 4
Ppcc 2
ppCC 1
ppCc 2
ppcc 1

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


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
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:

ZB

Zb

Zb

ZBZb

ZbZb

W

ZBW

ZbW

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.
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