Wrinkled, green, white
RRYY, RRYy, RRyy, RrYY, RrYy, Rryy, rrYY, rrYy, and rryy. There are three possible genotypes at each locus (when there are two alternative alleles); so the number of genotypes produced in the F2 of a cross between individuals heterozygous for n loci will be 3n. If there is incomplete dominance, the number of phenotypes also will be 3n because, with incomplete dominance, each genotype produces a different phenotype. If the traits exhibit dominance, the number of phenotypes will be 2n.
Not only are the principles of segregation and independent assortment important because they explain how heredity works, but they also provide the means for predicting the outcome of genetic crosses. This predictive power has made genetics a powerful tool in agriculture and other fields, and the ability to apply the principles of heredity is an important skill for all students of genetics. Practice with genetic problems is essential for mastering the basic principles of heredity — no amount of reading and memorization can substitute for the experience gained by deriving solutions to specific problems in genetics.
Students may have difficulty with genetics problems when they are unsure where to begin or how to organize the problem and plan a solution. In genetics, every problem is different, so there is no common series of steps that can be applied to all genetics problems. One must use logic and common sense to analyze a problem and arrive at a solution. Nevertheless, certain steps can facilitate the process, and solving the following problem will serve to illustrate these steps.
In mice, black coat color (B) is dominant over brown (b), and a solid pattern (S) is dominant over white spotted (s). Color and spotting are controlled by genes that assort independently. A homozygous black, spotted mouse is crossed with a homozygous brown, solid mouse. All the Fj mice are black and solid. A testcross is then carried out by mating the Fj mice with brown, spotted mice.
(a) Give the genotypes of the parents and the Fj mice.
(b) Give the genotypes and phenotypes, along with their expected ratios, of the progeny expected from the testcross.
Step 1: Determine the questions to be answered. What question or questions is the problem asking? Is it asking for genotypes, genotypic ratios, or phenotypic ratios? This problem asks you to provide the genotypes of the parents and the Fj, the expected genotypes and phenotypes of the progeny of the testcross, and their expected proportions.
Step 2: Write down the basic information given in the problem. This problem provides important information about the dominance relations of the characters and about the mice being crossed. Black is dominant over brown and solid is dominant over white spotted. Furthermore, the genes for the two characters assort independently. In this problem, symbols are provided for the different alleles (B for black, b for brown, S for solid, and s for spotted); had these symbols not been provided, you would need to choose symbols to represent these alleles. It is useful to record these symbols at the beginning of the solution:
B— black S— solid b — brown s—white-spotted
Next, write out the crosses given in the problem.
P homozygous X homozygous black, spotted brown, solid
Fj black, solid
Testcross black, solid X brown, spotted
Step 3: Write down any genetic information that can be determined from the phenotypes alone. From the pheno-types and the statement that they are homozygous, you know that the P-generation mice must be BBss and bbSS. The Fj mice are black and solid, both dominant traits, so the Fj mice must possess at least one black allele (B) and one spotted allele (S). At this point, you cannot be certain about the other alleles, so represent the genotype of the Fj as B?S?. The brown, spotted mice in the testcross must be bbss, because both brown and spotted are recessive traits that will be expressed only if two recessive alleles are present. Record these genotypes on the crosses that you wrote out in step 2:
P homozygous X homozygous black, spotted brown, solid BBss X bbSS
Fj black, solid
Testcross black, solid X brown, spotted B?S? X bbss
Step 4: Break down the problem into smaller parts.
First, determine the genotype of the Fj. After this genotype has been determined, you can predict the results of the testcross and determine the genotypes and pheno-types of the progeny from the testcross. Second, because this cross includes two independently assorting loci, it can be conveniently broken down into two single-locus crosses: one for coat color and another for spotting.
Third, use a branch diagram to determine the proportion of progeny of the testcross with different combinations of the two traits.
Step 5: Work the different parts of problem. Start by determining the genotype of the F1 progeny. Mendel's first law indicates that the two alleles at a locus separate, one going into each gamete. Thus, the gametes produced by the black, spotted parent contain Bs and the gametes produced by the brown, spotted parent contain bS, which combine to produce F1 progeny with the genotype BbSs:
P homozygous X homozygous black, spotted brown, solid BBss X bbss
1/2 ss spotted-* bbss brown, spotted
Step 6: Check all work. As a last step, reread the problem, checking to see if your answers are consistent with the information provided. You have used the genotypes BBss and bbSS in the P generation. Do these genotypes code for the phenotypes given in the problem? Are the F1 progeny phenotypes consistent with the genotypes that you assigned? The answers are consistent with the information.
Use the F1 genotype to work the testcross (BbSs X bbss), breaking it into two single-locus crosses. First, consider the cross for coat color: Bb X bb. Any cross between a heterozygote and a homozygous recessive genotype produces a 1:1 phenotypic ratio of progeny (see Table 3.2):
BB X bb
1/2 Bb black 1/2 bb brown
Next do the cross for spotting: Ss X ss. This cross also is between a heterozygote and a homozygous recessive genotype and will produce 1/2 solid (Ss) and 1/2 spotted (ss) progeny (see Table 3.2).
Ss X ss
1/2 Ss solid 1/2 ss spotted
Finally, determine the proportions of progeny with combinations of these characters by using the branch diagram.
1/2 Bb black * 1/2 X 1/2 = 1/4 1/2 ss spotted-* Bbss black, spotted y2 x y2 = V4
When two individuals of known genotype are crossed, we expect certain ratios of genotypes and phenotypes in the progeny; these expected ratios are based on the Mendelian principles of segregation, independent assortment, and dominance. The ratios of genotypes and phenotypes actually observed among the progeny, however, may deviate from these expectations.
For example, in German cockroaches, brown body color (Y) is dominant over yellow body color (y). If we cross a brown, heterozygous cockroach (Yy) with a yellow cockroach (yy), we expect a 1:1 ratio of brown (Yy) and yellow (yy) progeny. Among 40 progeny, we would therefore expect to see 20 brown and 20 yellow offspring. However, the observed numbers might deviate from these expected values; we might in fact see 22 brown and 18 yellow progeny.
Chance plays a critical role in genetic crosses, just as it does in flipping a coin. When you flip a coin, you expect a 1:1 ratio —1/2 heads and 1/2 tails. If you flip a coin 1000 times, the proportion of heads and tails obtained would probably be very close to that expected 1:1 ratio. However, if you flip the coin 10 times, the ratio of heads to tails might be quite different from 1:1. You could easily get 6 heads and 4 tails, or 3 and 7 tails, just by chance. It is possible that you might even get 10 heads and 0 tails. The same thing happens in genetic crosses. We may expect 20 brown and 20 yellow cockroaches, but 22 brown and 18 yellow progeny could arise as a result of chance.
If you expected a 1:1 ratio of brown and yellow cockroaches but the cross produced 22 brown and 18 yellow, you probably wouldn't be too surprised even though it wasn't a perfect 1:1 ratio. In this case, it seems reasonable to assume that chance produced the deviation between the expected and the observed results. But, if you observed 25 brown and 15 yellow, would the ratio still be 1:1? Something other than chance might have caused the deviation. Perhaps the inheritance of this character is more complicated than was assumed or perhaps some of the yellow progeny died before they were counted. Clearly, we need some means of evaluating how likely it is that chance is responsible for the deviation between the observed and the expected numbers.
To evaluate the role of chance in producing deviations between observed and expected values, a statistical test called the goodness-of-fit chi-square test is used. This test provides information about how well observed values fit expected values. Before we learn how to calculate the chi square, it is important to understand what this test does and does not indicate about a genetic cross.
The chi-square test cannot tell us whether a genetic cross has been correctly carried out, whether the results are correct, or whether we have chosen the correct genetic explanation for the results. What it does indicate is the probability that the difference between the observed and the expected values is due to chance. In other words, it indicates the likelihood that chance alone could produce the deviation between the expected and the observed values.
If we expected 20 brown and 20 yellow progeny from a genetic cross, the chi-square test gives the probability that we might observe 25 brown and 15 yellow progeny simply owing to chance deviations from the expected 20:20 ratio. When the probability calculated from the chi-square test is high, we assume that chance alone produced the difference. When the probability is low, we assume that some factor other than chance — some significant factor—produced the deviation.
To use the goodness-of-fit chi-square test, we first determine the expected results. The chi-square test must always be applied to numbers of progeny, not to proportions or percentages. Let's consider a locus for coat color in domestic cats, for which black color (B) is dominant over gray (b). If we crossed two heterozygous black cats (Bb X Bb), we would expect a 3:1 ratio of black and gray kittens. A series of such crosses yields a total of 50 kittens — 30 black and 20 gray. These numbers are our observed values. We can obtain the expected numbers by multiplying the expected proportions by the total number of observed progeny. In this case, the expected number of black kittens is 3/4 X 50 = 37.5 and the expected number of gray kittens is 1/4 X 50 = 12.5. The chi-square (x2) value is calculated by using the following formula:
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