ssyy

SsYy o

Gametes

10.7 Independent Assortment The 16 possible combinations of gametes in this dihybrid cross result in 9 different genotypes. Because Sand Yare dominant over s and y, respectively, the 9 genotypes result in 4 phenotypes in a ratio of 9:3:3:1.These results show that the two genes segregate independently.

10.7 Independent Assortment The 16 possible combinations of gametes in this dihybrid cross result in 9 different genotypes. Because Sand Yare dominant over s and y, respectively, the 9 genotypes result in 4 phenotypes in a ratio of 9:3:3:1.These results show that the two genes segregate independently.

Diploid parent

SsYy

Diploid parent

SsYy

Four haploid gametes

Four haploid gametes

/y/71 10.8 Meiosis Accounts for Independent Assortment of 11/ "h Alleles We now know that alleles of different genes are seg-\ / regated independently during metaphase I of meiosis.Thus a < ' parent of genotype SsYy can form gametes with four different genotypes.

on the same chromosome, as we will see below. However, it is correct to say that chromosomes segregate independently during the formation of gametes, and so do any two genes on separate homologous chromosome pairs (Figure 10.8).

One of Mendel's major contributions to the science of genetics was his use of the rules of statistics and probability to analyze his masses of data from hundreds of crosses producing thousands of plants. His mathematical analyses led to clear patterns in the data, and then to his hypotheses. Ever since Mendel, geneticists have used simple mathematics in the same ways that Mendel did.

Punnett squares or probability calculations: A choice of methods

Punnett squares provide one way of solving problems in genetics, and probability calculations provide another. Many people find it easiest to use the principles of probability, perhaps because they are so familiar. When we flip a coin, the law of probability states that it has an equal probability of landing "heads" or "tails." For any given toss of a fair coin, the probability of heads is independent of what happened in all the previous tosses. A run of ten straight heads implies nothing about the next toss. No "law of averages" increases the likelihood that the next toss will come up tails, and no "momentum" makes an eleventh occurrence of heads any more likely. On the eleventh toss, the odds of getting heads are still 50/50.

The basic conventions of probability are simple:

► If an event is absolutely certain to happen, its probability is 1.

► If it cannot possibly happen, its probability is 0.

► Otherwise, its probability lies between 0 and 1.

A coin toss results in heads approximately half the time, so the probability of heads is xh—as is the probability of tails.

multiplying probabilities. How can we determine the probability of two independent events happening together? If two coins (a penny and a dime, say) are tossed, each acts independently of the other. What, then, is the probability of both coins coming up heads? Half the time, the penny comes up heads; of that fraction, half the time the dime also comes up heads. Therefore, the joint probability of both coins coming up heads is half of one-half, or V2 x 1/2 = 1/4. To find the joint probability of independent events, then, we multiply the probabilities of the individual events (Figure 10.9). How does this method apply to genetics?

the monohybrid cross. To apply the principles of probability to genetics problems, we need only deal with gamete formation and random fertilization instead of coin tosses. A homozygote can produce only one type of gamete, so, for example, the probability of an SS individual producing gametes with the genotype S is 1. The heterozygote Ss produces S gametes with a probability of 1/2, and s gametes with a probability of V2.

Consider the F2 progeny of the cross in Figure 10.4. They are obtained by self-pollination of F1 plants of genotype Ss. The probability that an F2 plant will have the genotype SS must be V2 x V2 = M, because there is a 50:50 chance that the sperm will have the genotype S, and that chance is independent of the 50:50 chance that the egg will have the genotype S. Similarly, the probability of ss offspring is V2 x V2 = M.

adding probabilities. How are probabilities calculated when an event can happen in different ways? The probability of an F2 plant getting an S allele from the sperm and an s allele from the egg is M, but remember that the same

Two coin tosses are independent events with an outcome probability (P) of V2 each.

Two coin tosses are independent events with an outcome probability (P) of V2 each.

There are two ways to arrive at a heterozygote, so we add the probabilities of the two individual outcomes:

There are two ways to arrive at a heterozygote, so we add the probabilities of the two individual outcomes:

10.9 Using Probability Calculations in Genetics The probability of any given combination of alleles from a sperm and an egg appearing in the offspring of a cross can be obtained by multiplying the probabilities of each event. Since a heterozygote can be formed in two ways, these two probabilities are added together.

genotype can also result from an s from the sperm and an S from the egg, also with a probability of M. The probability of an event that can occur in two or more different ways is the sum of the individual probabilities of those ways. Thus the probability that an F2 plant will be a heterozygote is equal to the sum of the probabilities of the two ways of forming a heterozygote: Vi + M = V2 (see Figure 10.9). The three genotypes are therefore expected in the ratio V4 SS : V2 Ss : Vi ss—hence the 1:2:1 ratio of genotypes and the 3:1 ratio of phenotypes seen in Figure 10.4.

the dihybrid cross. If F1 plants heterozygous for two independent characters self-pollinate, the resulting F2 plants express four different phenotypes. The proportions of these phenotypes are easily determined by probability calculations. Let's see how this works for the experiment shown in Figure 10.7.

Using the principle described above, we can calculate that the probability that an F2 seed will be spherical is 3/4: the probability of an Ss heterozygote (V2) plus the probability of an SS homozygote (M) = ^ By the same reasoning, the probability that a seed will be yellow is also ^ The two characters are determined by separate genes and are independent of each other, so the joint probability that a seed will be both spherical and yellow is ^ x ^ = 9/16. What is the probability of F2 seeds being both wrinkled and yellow? The probability of being yellow is again the probability of being wrinkled is M> x 1/2 = M. The joint probability that a seed will be both wrinkled and yellow, then, is 1/4 x 3/4 = //16. The same probability applies, for similar reasons, to spherical, green F2 seeds. Finally, the probability that F2 seeds will be both wrinkled and green is M x M = Vi6. Looking at all four phenotypes, we see they are expected in the ratio of 9:3:3:1.

Probability calculations and Punnett squares give the same results. Learn to do genetics problems both ways, and then decide which method you prefer.

Mendel's laws can be observed in human pedigrees

After Mendel's work was uncovered by plant breeders, Mendelian inheritance was observed in humans. Currently, the patterns of over 2,500 inherited human characteristics have been described.

How can Mendel's laws of inheritance be applied to humans? Mendel worked out his laws by performing many planned crosses and counting many offspring. Neither of these approaches is possible with humans. So human geneticists rely on pedigrees, family trees that show the occurrence of phenotypes (and alleles) in several generations of related individuals.

Because humans have such small numbers of offspring, human pedigrees do not show the clear proportions of offspring phenotypes that Mendel saw in his pea plants (see Table 10.1). For example, when two people who are both heterozygous for a recessive allele (say, Aa) marry, there will be, for each of their children, a 25 percent probability that the child will be a recessive homozygote (aa). Thus, over many such marriages, one-fourth of all the children will be recessive homozygotes (aa). But the offspring of a single marriage are likely to be too few to show the exact one-fourth proportion. In a family with only two children, for example, both could easily be aa (or Aa, or AA).

To deal with this ambiguity, human geneticists assume that any allele that causes an abnormal phenotype is rare in the human population. This means that if some members of a given family have a rare allele, it is highly unlikely that an outsider marrying into that family will have that same rare allele.

Human geneticists may wish to know whether a particular rare allele is dominant or recessive. Figure 10.10 depicts a pedigree showing the pattern of inheritance of a rare domi-

Generation I (parents)

Generation II

Generation III

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