Chapter 2 -- Mendel's Principles

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

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.

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

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".

RR: red flower TT: tall plant Rr: pink flower Tt: medium height plant rr: white flower tt: dwarf plantIf 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

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.

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.

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.

a. Type A

In this case a type A person would have either the genotype

In this case a type B person would have either the genotype

In this case a type O person would have the genotype

In this case a type AB person would have the genotype

Thus, in this case none of the men can be excluded from possible paternity.

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 = 1024If 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].

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.

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).

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. = 1This 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.

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.

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

b) short, green, wrinkled

c) short, green, smooth

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*

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

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.

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

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.

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

Take the ABO locus first. genotype of the dean A- genotype of Steve's father A- genotype of the dean's daughter OOIn 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 bloodIn 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 SSIn 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.