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# Patterns of Inheritance Practice Problems (page 4)

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By — McGraw-Hill Professional
Updated on Aug 19, 2011

#### Practice 19

In the garden pea, yellow cotyledon color is dominant to green, and inflated pod shape is dominant to the constricted form. When both of these traits were considered jointly in self-fertilized dihybrids, the progeny appeared in the following numbers : 193 green, inflated : 184 yellow, constricted : 556 yellow, inflated : 61 green, constricted. Test the data for independent assortment.

#### Solution 19

This is not a significant chi-square value, and thus we accept the null hypothesis, i.e., the magnitude of the deviation (oe) is to be expected by chance alone in greater than 95% of an infinite number of experiments of comparable size. This is far above the critical value of5%necessary for acceptance of the hypothesis. We may therefore accept the data as being in conformity with a 9 : 3 : 3 : 1 ratio, indicating that the gene for cotyledon color assorts independently of the gene for pod form.

#### Practice 20

A total of 160 families with 4 children each were surveyed with the following results:

Is the family distribution consistent with the hypothesis of equal numbers of boys and girls?

#### Solution 20

This value is close to, but less than, the critical value 9.49.We may therefore accept the hypothesis, but the test would be more definitive if it could be run on a larger sample. It is a well-known fact that a greater mortality occurs in males than in females and therefore an attempt should be made to ascertain family composition on the basis of sex of all children at birth, including prematures, aborted fetuses, etc. In the U.S., the natural boy/girl ratio is approximately 105/100. To simplify calculations in this book, assume a ratio of 1.0 unless otherwise stated.

#### Practice 21

In Problem 2.18 (b), it was shown that observations of 60 : 40 produced a significant chi-square at the 5% level when uncorrected for continuity. Apply the Yates correction for continuity and retest the data.

#### Solution 21

Notice that the correction 0.5 is always applied to the absolute value |oe| of the deviation of expected from observed numbers. This is not a significant chi-square value. Because the data are discrete (jumping from unit to unit) there is a tendency to underestimate the probability, causing too many rejections of the null hypothesis. The Yates correction removes this bias and produces a more accurate test near the critical values (column headed by a probability of 0.05 in Table 2-3).

#### Practice 22

The black hair of guinea pigs is produced by a dominant gene B and white by its recessive allele b. Unless there is evidence to the contrary, assume that II1 and II4 do not carry the recessive allele. Calculate the probability that an offspring of III1 × III2 will have white hair.

#### Solution 22

Both I1 and I2 must be heterozygous (Bb) in order to have the white (bb) offspring II2. If III1 or III2 had been white, this would constitute evidence that II1 or II4 were heterozygous. In the absence of this evidence, the problem tells us to assume that II1 and II4 are homozygous (BB). If the offspring of III1 × III2 is to be white, then both III1 and III2 would have to be heterozygous (Bb). In this case, II3 would also have to be heterozygous in order to pass the recessive allele on to III2.Under the conditions of the problem, we are certain that III1 is heterozygous because his parents (II1 × II2) are BB × bb. We notice that II3 is black. The probability that black progeny from I1 × I2 are heterozygous is 2=3. If II3 is heterozygous, the probability that III2 is heterozygous is 1/2. If III2 is heterozygous, there is a 25% chance that the offspring of IIIl × III2 will be white (bb). Thus, the combined probability that II3 is heterozygous and III2 is heterozygous and producing a white offspring is the product of the independent probabilities = (2/3) (1/2) (1/4) = 2/24 = 1/12.

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