PROBLEM SET
Chapter 2 -- Mendel's Principles

P. 39, #2. Mendel self-fertilized dihybrid plants (RrYy) with round and yellow seeds and got a 9:3:3:1 ratio in the F2 generation. As a test of Mendel's hypothesis of independent assortment, predict the kinds and numbers of progeny produced in testcrosses of these F2 offspring.
You need to set up testcrosses with each of the possible phenotypes that are present in the F2 generation, however, these four phenotypes represent 9 genotypes, so we need to predict the outcome of 9 different crosses. A testcross is an individual mated with an individual that is homozygous recessive at every locus, in this case rryy. The crosses are listed below. Each of the 9 genotypes of the F2 generation, yields a unique result when testcrossed, thus each genotype of the F2 can be unequivocally determined by this method.

ROUND, YELLOW (R-Y-)
   1/16 RRYY x rryy -----> all RrYy  round, yellow

   2/16 RrYY x rryy -----> 1/2 RrYy  round, yellow
                           1/2 rrYy  wrinkled, yellow

   2/16 RRYy x rryy -----> 1/2 RrYy  round, yellow
                           1/2 Rryy  round, green

   4/16 RrYy x rryy -----> 1/4 RrYy  round, yellow
                           1/4 rrYy  wrinkled, yellow
                           1/4 Rryy  round, green
                           1/4 rryy  wrinkled, green

WRINKLED, YELLOW (rrY-)
   1/16 rrYY x rryy -----> all rrYy  wrinkled, yellow

   2/16 rrYy x rryy -----> 1/2 rrYy  wrinkled, yellow
                           1/2 rryy  wrinkled, green

ROUND, GREEN (R-yy)
   1/16 RRyy x rryy -----> all Rryy  round, green

   2/16 Rryy x rryy -----> 1/2 Rryy  round, green
                           1/2 rryy  wrinkled, green

WRINKLED, GREEN (rryy)
   1/16 rryy x rryy -----> all rryy  wrinkled, green
P. 39, #3. In a variety of onions three bulb colors segregate: red, yellow, and white. A plant with a red bulb is crossed to a plant with a white bulb and the offspring have all red bulbs. When these are selfed the following plants are obtained:
red-bulbed 119
yellow-bulbed 32
white-bulbed 9
What is the mode of inheritance of bulb color and how do you account for the ratio?

Because you get three phenotypes that are clearly not in a 1:2:1 ratio as you would expect from a monohybrid cross with incomplete dominance, then you must assume that it is some kind of dihybrid cross. You then need to determine the phenotypic ratio based upon the assumption of two loci. Add the numbers of the three phenotypes to get the total number of offspring, then divide each number of offspring by the total, and finally multiply by 16 (the number of gametic genotypes squared or the number of cells in the Punnett square). This will give you the ratio of three phenotypes to each other.
red-bulbed (119/160)*16 = 11.9 approx. 12
yellow-bulbed (32/160)*16 = 3.2 approx. 3
white-bulbed (9/160)*16 = 0.9 approx. 1
Thus the approximate phenotypic ratio is 12:3:1, which is some permutation of the standard dihybrid ratio of 9:3:3:1 with two of the phenotypic classes combined. Thus we should think of some kind of epistatic process. The phenotypic ratios are such that
9/16 R-Y- red
3/16 R-yy red
3/16 rrY- yellow
1/16 rryy white
The red phenotype occurs whenever there is one dominant allele at the "R locus" and the genotype at the "Y locus" has no effect on the phenotype, thus the "R locus" is epistatic to the "Y locus" when the "R allele" is present at the "R locus".

P. 40, #4. Four o'clock plants have a gene for color and a gene for height with the following phenotypes.
  RR: red flower            TT: tall plant
  Rr: pink flower           Tt: medium height plant
  rr: white flower          tt: dwarf plant
If a dihybrid plant is self-fertilized, give the proportions of genotypes and phenotypes produced.

The dihybrid plant is heterozygous at both loci, thus its genotype is RrTt (its phenotype is pink-flowered and medium height). To self it is the same as the cross RrTt x RrTt. From the rule of independent assortment, we know that each dihybrid will produce four types of gametes (gametic genotypes) in equal porportions to each other. The Punnett square for the cross is below.
An examination of the Punnett square shows 9 different genotypes in the ratio given below. Because the expression of the alleles is an example of incomplete dominance, there are 9 different phenotypes with each matching to one genotype.








number   genotype          phenotype

  1        RRTT     red-flower, tall plant
  2        RrTT     pink flower, tall plant
  1        rrTT     white flower, tall plant
  2        RRTt     red flower, medium height plant
  4        RrTt     pink flower, medium height plant
  2        rrTt     white flower, medium height plant
  1        RRtt     red flower, dwarf plant
  2        Rrtt     pink flower, dwarf plant
  1        rrtt     white flower. dwarf plant
P. 40, #6. When studying an inherited phenomenon, a geneticist discovers a phenotypic ratio of 9:6:1 among offspring of a given mating. Give a simple, plausible explanation of the results. How would you test this hypothesis?
As 9:6:1 appears to be a variant of the standard 9:3:3:1 ratio you would expect from a dihybrid cross, the simplest explanation is that this result is from a dihybrid cross in which epistasis plays a role. In this case, you would expect the phenotypes to have the have the genotypes given below.
9/16 A-B-
6/16 A-bb and aaB-
1/16 aabb
To test this, you would perform a series of testcrosses that would be similar to those done in problem #2 and see if the results of your crosses match your predictions.

P.40, #7. In order to determine the genotypes of the offspring of a cross in which a corn trihybrid (AaBbCc) was selfed, a geneticist has three choices. He or she can take a sample of the progeny and (a) self-fertilize the individual plants, (b) testcross the plants, or (c) cross the individuals with a trihybrid (backcross). Which method is preferred?
When doing a cross to determine the genotype of an individual, the object is to determine if the individual with a dominant phenotype is heterozygous or homozygous at the locus in question. The problem is to reveal the recessive allele, if it is present. So you want a method that will most reliably reveal the recessive allele. The preferred method in all cases in method (b) the testcross. The reason is that in a testcross, half of all offspring should be homozygous recessive and thus half of all the offspring will have the recessive phenotype. With the other methods only one quarter of the offspring will have a homozygous recessive genotype and thus express the recessive phenotype. As there is twice as many recessive individuals with a testcross, you have a much greater chance of revealing the recessive alleles in the trihybrid. This becomes important when doing crosses that result in few numbers of offspring, for example Labrador Retrievers.

P. 40, #9. In the ABO blood system in human beings, alleles A and B are codominant and both are dominant to the O allele. In a paternity dispute, a type AB woman claimed that one of four men was the father of her type A child. Which of the following men could be the father of the child on the basis of the evidence given?
a. Type A

In this case a type A person would have either the genotype AA or AO. A man with either of these genotypes could be the father as the mother would donate the A allele to the child and either an A allele from the father or an O allele from the father would produce a child with Type A blood.
b. Type B
In this case a type B person would have either the genotype BB or BO. A man with the genotype BO could be the father as the mother would donate the A allele to the child and an O allele from the father would produce a child with Type A blood.
c. Type O
In this case a type O person would have the genotype OO. A man this genotype could be the father as the mother would donate the A allele to the child and an O allele from the father would produce a child with Type A blood.
d. Type AB
In this case a type AB person would have the genotype AB. A man with this genotype could be the father as the mother would donate the A allele to the child and an A allele from the father would produce a child with Type A blood.
Thus, in this case none of the men can be excluded from possible paternity.

P. 40, #14. A geneticist crossed two corn plants, creating an F1 decahybrid (ten segregating loci). This decahybrid was then self-fertilized. How many different kinds of gametes were produced by the F1 plant? What proportion of the F2 offspring were recessive homozygotes? How many different kinds of genotypes and phenotypes were generated in the F2 offspring? What would your answers be if the decahybrid were testcrossed instead?
You need to use the chart that was given in lecture to answer these questions. If you try to work this problem by diagraming the cross, it will be intractable because the Punnett Square would be 1024 rows by 1024 columns. The chart is given below with the answers inserted in the appropriate place (2^n means 2 raised to the nth power; 2^3 = 2x2x2 = 8). Remember that a decahybrid is heterozygous at 10 loci (AaBbCcDdEeFfGgHhIiJj).
                       mono   tri        deca        
# of gametes             2     8       2^10 = 1024  

prop. of
recessive homozygotes   1/4  1/64   (1/(2^10))^2 = 1/1048576
in the F2 generation

# of genotypes in
the F2 generation       3     27       3^10 = 59049

# of phenotypes in
the F2 generation       2      8       2^10 = 1024
If the decahybrid was testcrossed the only numbers that would change would be the prop. of recessive homozygotes [(1/(2^10)) = 1/1024] and the number of genotypes in the F2 generation [(2^10) = 1024].

P.41, #27. Maple sugar urine disease is a rare inborn error of human metabolism. The urine of affected individuals smells like maple sugar.
a) If two unaffected individuals have an affected child, what is the mode of inheritance of the disease?

As you get an affected child from unaffected parents, you have to assume that the maple sugar urine disease is a recessive trait and that both parents are heterozygous for the disease and that the affected child is homozygous. You assume that the trait is on an autosome because neither the mother nor the father expresses the trait. If it was on the X-chromosome, the father would express the trait because he would be hemizygous and not have another allele to mask the recessive allele. Thus the mode of inheritance is autosomal recessive.
b) What is the chance that the second child will be unaffected?
The cross is basically that of a monohybrid cross. In a monohybrid cross the dominant phenotype (which is "unaffected" in this case) occurs in three out of four offspring. The phenotype of the first child has no effect on the phenotype of the second child. Thus the probability of a second child being unaffected is 3/4 (0.75).

P. 42, #33. A plant with red flowers is crossed to a plant with white flowers. All the progeny are pink. When the pink flowers are crossed, the progeny are 11 red, 23 pink, and 12 white. What is the mode of inheritance of color?
First, the pink flower in the F1 generation must be heterozygous (Rr), and second, the phenotypic ratio does seem to be close to 1:2:1 and thus you can assume that only one locus is involved. To determine precisely what the phenotypic ratio is, add the numbers of the three phenotypes to get the total number of offspring, then divide each number of offspring by the total, and finally multiply by 4 (the number of gametic genotypes squared or the number of cells in the Punnett square).
    red     (11/46)*4 = 0.96  approx. = 1
    pink    (23/46)*4 = 2.00  approx. = 2
    white   (12/46)*4 = 1.04  approx. = 1
This is the genotypic ratio that you expect in the F2 generation of a monohybrid cross. Because the phenotypic ratio matches the genotypic ratio, the red allele must be incompletely dominant to the white allele.

P. 42, #36 A brown-eyed, long-winged fly is mated to a red-eyed, long-winged fly. The progeny are:
51 long, red
53 long, brown
18 short, red
16 short, brown
What are the genotypes of the parent?

In this case, it is easier to look at each locus separately. At the wing locus, we have two long-winged flies crossed to yield 104 long-winged flies and 34 short-winged flies. This is very close to a 3:1 ratio that we would expect from a monohybrid cross. Thus, the parents must be heterozygous (Ll) at the wing-length locus and long wings must be dominant. At the eye color locus, we have a red-eyed fly crossed with a brown-eyed fly to yield 69 brown-eyed flies and 69 red-eyed flies. This is a 1:1 ratio, which is what we would expect from a monohybrid testcross. However, we do not know which is dominant, red eyes or brown eyes. Thus one parent is heterozygous (Rr) and the other parent is homozygous recessive (rr) at the eye color locus. Combining the information from the two loci, possible genotypes for the parents are LlRr for the brown-eyed, long-winged parent and Llrr for the red-eyed, long-winged parent. The other possibility is Llrr for brown-eyed, long-winged and LlRr for red-eyed, long-winged.

P. 42, #38. In peas, tall (T) is dominant to short (t), yellow (Y) is dominant to green (y), and smooth (S) is dominant to wrinkled (s). From a cross of two triple heterozygotes, what is the chance of getting a plant that is:
a. tall, yellow, smooth
b. short, green, wrinkled
c. short, green smooth

This is a standard trihybrid cross with complete dominance. The cross is a simple extension of a dihybrid cross and could be worked by diagraming the cross with a Punnett Square. However, each trihybrid will produce 8 different gametes and thus the Punnett Square will be an 8x8 square (64 cells). This would be rather tedious to draw and an easier way is to consider the probabilities of occurrence of each of the phenotypes and then to calculate the probability of these various combinations. Because of the Rule of Segregation we know that in the results of a monohybrid cross, the probability of an individual having the dominant phenotype is 0.75 and the probability of an individual having the recessive phenotype is 0.25. We know from the Rule of Independent Assortment that the presence of other loci in the cross does not affect these probabilities. Thus for the loci, the probability of being yellow is 0.75 and green is 0.25, the probability of being tall is 0.75 and short is 0.25, and the probability of being smooth is 0.75 and wrinkled is 0.25. From these known probabilities, the probability of each phenotypic combination can be calculated from the product rule.
a) tall, yellow, smooth
prob(tall) x prob(yellow) x prob(smooth) =
0.75 x 0.75 x 0.75 = 0.421875
b) short, green, wrinkled
prob(short) x prob(green) x prob(wrinkled) =
0.25 x 0.25 x 0.25 = 0.015625
c) short, green, smooth
prob(short) x prob(green) x prob(smooth)
0.25 x 0.25 x 0.75 = 0.046875

P. 42, #42 A, B, and C are independently assorting Mendelian factors controlling the production of black pigment; alleles of these genes are indicated a, b, and c respectively. Assume that A, B, and C act in a pathway as follows:
A black AABBCC individual is crossed with a colorless aabbcc to give black F1 individuals. The F1 individuals are selfed to give F2 progeny. a) What proportion of the F2 generation is colorless? b) What proportion of the F2 generation is red?
There are 27 different genotypes possible in the F2 generation. These are as follows:
  1/64 AABBCC         2/64 AABBCc          1/64 AABBcc+
  2/64 AaBBCC         4/64 AaBBCc          2/64 AaBBcc+
  1/64 aaBBCC+        2/64 aaBBCc+         1/64 aaBBcc+
  2/64 AABbCC         4/64 AABbCc          2/64 AABbcc+
  4/64 AaBbCC         8/64 AaBbCc          4/64 AaBbcc+
  2/64 aaBbCC+        4/64 aaBbCc+         2/64 aaBbcc+
  1/64 AAbbCC+        2/64 AAbbCc+         1/64 AAbbcc+
  2/64 AabbCC+        4/64 AabbCc+         2/64 Aabbcc+
  1/64 aabbCC*        2/64 aabbCc*         1/64 aabbcc*
a) What proportion of the F2 generation is colorless?
For an individual to be colorless it must not be able to produce the red pigment. There are two ways that it can produce the red pigments, by either an "A allele" at the "A locus" or by a "B allele" and the "B locus". Thus for an individual to be colorless, that individual must be homozygous recessive at both the "A locus" and the "B locus". The alleles at the "C locus" will have no effect on the phenotype. The genotype could be written as aabb--. These are on the bottom row of the table above and indicated with asterisks *. The proportion of colorless individual is thus 1/64 + 2/64 + 1/64 = 4/64 (or 1/16).
Alternatively, you could solve this using the rules of probability. The probability of being homozygous aa at the "A locus" is 1/4 and the probability of being bb at the "B locus" is also 1/4. The probability of being both aa and bb is 1/4 x 1/4 or 1/16, which is the proportion of colorless that we determined above.

b) What proportion of the F2 generation is red?
To be red, an individual must be able to produce either of the red pigments and not be able to convert the red pigments to black. To not be able to convert to black, the individual can be homozygous cc at the "C locus" or be missing one of the red pigments (it needs both to make the black pigment). There are several ways in which this can occur. The genotypes are indicated with a + in the above table.
A-B-cc    at least one dominant allele at both the "A locus" and
          the "B locus" and homozygous recessive at the "C
          locus".
A-bbcc    at least one dominant allele at the "A locus" and
          homozygous recessive at the "B locus" and "C locus".
aaB-cc    at least one dominant allele at the "B locus" and
          homozygous recessive at the "A locus" and the "C locus"
aaB-C-    at least one dominant allele at both the "B locus" and
          the "C locus" and homozygous recessive at the "A locus"
A-bbC-    at least one dominant allele at both the "A locus" and
          the "C locus" and homozygous recessive at the "B locus"

The probability of occurrence of A-B-cc is the product of the
prob(A-) x prob(B-) x prob(cc) = 0.75 x 0.75 x 0.25 = 0.140625.
The probability of occurrence of A-bbcc is the product of the prob(A-) x prob(bb) x prob(cc) = 0.75 x 0.25 x 0.25 = 0.046875.
The probability of occurrence of aaB-cc is the product of the prob(aa) x prob(B-) x prob(cc) = 0.25 x 0.75 x 0.25 = 0.046875.
The probability of occurrence of aaB-C- is the product of the prob(aa) x prob(B-) x prob(C-) = 0.25 x 0.75 x 0.75 = 0.140625.
The probability of occurrence of A-bbC- is the product of the prob(A-) x prob(bb) x prob(C-) = 0.75 x 0.25 x 0.75 = 0.140625.
The sum of these five probabilities is 0.5156, which is equal to 33/64.


P. 43, #44. A pre-med student, Steve, plans to marry the daughter of the dean of nursing. Steve's father, a doctor, puts pressure on Steve to marry someone else because the girl has a sister with PKU (phenylketonuria) and a brother with albinism. The dean's husband was sterile and the three children were all the result of artificial insemination from three different donors. Having served as an anonymous sperm donor, the doctor is concerned that Steve and his fiance‚ may be half brother and sister. Given the following information, deduce whether Steve and his fiance are related. The MN and Ss systems are two independent, codominant blood genetic systems.
                    Blood type
------------------------------
dean                 A  MN  Ss
dean's daughter      O  M   S
Steve's father       A  MN  Ss
Steve                O  N   s
Steve's mother       B  N   s
We need to show that Steve's father can not be the father of the dean's daughter. If we can show this then Steve and the dean's daughter can not be half-brother and sister. We need to demonstrate the impossibility of paternity at only one locus of the three, thus the problem is easier if we take one locus at a time.
Take the ABO locus first.
    genotype of the dean A-
    genotype of Steve's father A-
    genotype of the dean's daughter OO
In this case if both the dean and Steve's father are heterozygous, then it is possible that Steve's father is also the father of the dean's daughter. As we do not know that the dean and Steve's father are not heterozygous, we can not rule out paternity based on this locus.
Take the MN locus next
     genotype of the dean             MN
     genotype of Steve's father       MN  type MN blood
     genotype of the dean's daughter  MM  type M blood
In this case the dean and Steve's father could have both donated "M alleles" to the dean's daughter. Thus, we can not rule out paternity based on this locus.
Take the Ss locus next
     genotype of the dean             Ss
     genotype of Steve's father       Ss
     genotype of the dean's daughter  SS
In this case the dean and Steve's father could have both donated "S alleles" to the dean's daughter. Thus, we can not rule out paternity based on this locus. We can not prove that Steve and the dean's daughter are not related and thus they could be half brother and sister.

Last updated on 24 September 1996.
Provide comments to Dwight Moore at mooredwi@esumail.emporia.edu.
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