Chi-Square Test and Probability

where d represents deviation from expected ratio, e the expected ratio, and Σ is the sum. The smaller the chi-square value the more likely it is that deviation has occurred due to chance.

TABLE 1. Chi-square analysis of F2 progenies of three crosses
between tall and dwarf pea plants

Probability:

In the experiments described above the F2 generation segregates in a particular ratio. The numbers of individuals in each ratio result from chance segregation of genes during gamete formation, and their chance combinations to form zygotes. Since these are chance events, accurate predictions about the results cannot be made.

This is especially true in cases where the progeny is limited to a small number such as in experiments in animal breeding and even more so in human pedigree studies. It is hardly possible to predict the appearance of a certain phenotype or genotype. But what we can say is that there is a certain probability of occurrence
of a given genetic event.

In a general way we can say that the probability or chance that an event will occur can be defined as the proportion of times in which that event occurs in a very large number of trials. If there are x trials and the event occurs on the average y times during the x trials, then the probability is expressed as y/x. The probability would be between zero and one.

The closer the probability is to zero, the less chance is there for the event to occur. When probability is one, the event is certain to occur.\

Rules of Probability:

The multiplication rule: While considering probabilities it is important to note that the inheritance of certain genes such as for height of pea plants (tall or dwarf), flower colour (red or white), seed texture (round or wrinkled) are independent events. If we consider each gene separately, the probability of any one F2 plant being tall is 3/4, and of its being dwarf 1/4.

Similarly there is a probability of 3/4 that an F2 plant will bear red flowers and 1/4 that the flower will be white. Now what is the probability that an F2 plant would be both tall and coloured? Assuming that the inheritance of each gene is an independent event, the probability that a plant be coloured and tall is equal to the product of their individual probabilities, i.e., 3/4 × 3/4 = 9/16.

Thus there are 9 chances out of 16 that an F2 plant be coloured and tall. Stated in a general way, when the probability of an event is independent of that of another event, and the occurrence of one does not influence the occurrence of the other, probability that both events will occur together is the product of their individual probabilities.

The Addition Rule: The probabilty that one of several mutually exclusive events will occur is the sum of their individual probabilities. This law is applicable when different types of events cannot occur together. If one occurs the other is excluded. When a coin is tossed there are two alternatives, either heads or tails will appear.

If the probability for heads is 1/2, for tails 1/2, then the probability that either heads or tails will appear is 1/2 + 1/2 = 1.

Probability and Human Genetics:

In some recessive genetic traits in humans such as phenylketonuria (PKU), albinism and others, the birth of an affected child indicates that the parents are heterozygous carries for that trait. Due to the recessive nature of the gene the parents are healthy and normal. Such parents could be anxious to know what chances exist for any of their future children to show the genetic defect.

Since the trait is recessive they could expect affected children and normal children in the ratio 1 : 3. But since human families are limited in size, this ratio can be misleading. All we can say is that the chance or probability of having an affected child is 1/4 at each birth. Moreover, even after the birth of an affected child, the same probability (1/4) exists for an affected child in all future pregnancies.

The probability for a normal child at each birth is 3/4. The rules of probability can be applied for predicting the ratio of boys and girls born in a family. Since the human male produces an equal number of X and Y sperm, the chance for a boy at any birth is 1/2, and for a girl also is 1/2. From the probability of each single conception it is possible to calculate the probability of successive births together.

For example, what is the probability that the first two children born in a family will both be males. To find this out we must determine the product of the separate probabilities at each conception, that is 1/2 × 1/2 = 1/4. Now consider a different question.

What is the probability that the third child in the family in which the first two are males, will also be a male ? For the answer we must remember that the sex of any child is independent of the sex of the other children; therefore the probability for the third child to be male is 1/2.

Binomial Expansions

There are many situations in genetics where we would like to know the probability that a combination of events will occur. For instance it may be required to determine the probability that two offsprings of a mating between Aa and aa parents will have a particular genetic constitution, namely 2 Aa, 2 aa, and 1 Aa and the other aa.
The occurrence of a particular genotype in a single offspring is an independent event, as it is not influenced by the genotype of any other offspring. The probability that 2 Aa offspring will be formed from this mating is therefore equal to the product of their separate probabilities.

Aa = 1/2 × 1/2 = 1/4 or 25%

Thus the probabilities for each sequence of two offspring”s are as follows:

Note that the probability for a complementary event such as the probability that both offspring will not be aa is 1 – the probability of the particular event, or 1 – 1/4 = 3/4. If the probabilities are calculated for the different combinations of genotypes possible among 3 children of the mating Aa × aa, the frequencies of each combination will be found to correspond with raising a binomial to the third power.

Probability that 3 offspring are Aa = 1/8
Probability that two are Aa and one aa = 3/8
Probability that two are aa and one Aa = 3/8
Probability that 3 offspring are aa = 1/8
or (p + q)3 = 1p3 + 3p2q + 3pq2 + 1q3
or [(Aa) + (aa)]3 = 1(Aa) (Aa) (Aa) + 3(Aa) (Aa) (aa) + 3(Aa) (aa) (aa) + 1(aa) (aa) (aa)

The probability for each particular combination of offspring can therefore be determined by the binomial coefficient for that combination relative to the total number of possible combinations. In general we can say that when p is the probability that a particular event will occur and q or 1 – p is the probability of an alternative form of that event so that p + q = 1, then probability for each combination in which a succession of such events may occur is described by the binomial distribution.

Multinomial Distribution:

In the examples described above there are only two possible alternatives for each event. Sometimes a genetic cross or mating can produce three types of offspring. For instance Aa × Aa will produce AA, Aa and aa offspring in the ratio 1 : 2 : 1 so that their probabilities are 1/4 for AA, 1/2 for Aa and 1/4 for aa.

In such cases an additional term is added to the binomial to represent the third class of offspring. We now have a trinomial distribution (p + q + r)n, where p, q and r represent probabilities of AA, Aa and aa respectively.

For determining probabilities of trinomial combinations we use the formula

where w, x and y are the numbers of offspring of each of the three different types, and p, q and r are their probabilities respectively. In a mating of Aa × Aa where only four offspring are produced, the probability of having exactly 1 AA homozygote, 2 Aa heterozygote and 1 aa homozygote would be

Other multinomial distributions can similarly be worked out. For example when a dihybrid cross is made in which four phenotypes can appear with frequencies p, q, r and s, the formula would be
As an illustration, consider the cross between smooth and yellow pea plants with wrinkled green plants described in Chapter 1. When the F1 heterozygote is self-fertilised, 19 offspring are obtained of which 8 are smooth yellow (probability 9/16), 5 are smooth green (probability 3/16), 4 wrinkled yellow (probability 3/16) and 2 wrinkled green (probability 1/16). The probability of having the 19 offspring in exactly this ratio (9 : 3 : 3 : 1) would be

Questions:

  1. In a plant heterozygous for striped leaves (Ss) what is the probability that a pollen grain will receive the S gene?

  2. In cattle hornless condition is dominant over horned. What is the probability that the matings of two heterozygous cattle should result in:
    (a) a calf with horns;
    (b) a hornless calf.

  3. At each conception the probability for a child being male is 1/2, and of female also 1/2. If a couple have three girls, what is the probability that their fourth child would be a boy?

  4. In a plant heterozygous for purple flower colour and round seed (PpRr) what is the probability that a pollen grain will contain:
    (a) the P allele;
    (b) P and R alleles;
    (c) P and r alleles;
    (d) the r allele.

  5. Drosophila has 4 chromosomes. Considering any three genes at random, they will have equal chance to be located on any of the four chromosomes. What is the probability that all three should be present on the same chromosome?

References:

  • Bailey, N.T.J. 1959. Statistical Methods in Biology. John Wiley, New York.
  • Dunn, O.J. 1964. Basic Statistics, A Primer for the Biomedical Sciences. John Wiley, New York.
  • Kempthorne, O. 1957. An Introduction to Genetic Statistics. John Wiley, New York.
  • Strickberger, M.W. 1976. Genetics. Macmillian, New York.
  • Winchester, A.M. 1977. Genetics. Houghton, Mifflin, Boston.
  • Fisher, R. A. and Yates, F. 1963. Statistical Tables for Biological, Agricultural and Medical Research, 6th ed. Oliver and Boyd, Edinburgh.

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