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# Patterns of Inheritance Practice Problems (page 3)

based on 2 ratings
By McGraw-Hill Professional
Updated on Aug 19, 2011

#### Practice 13

In the garden pea, Mendel found that yellow seed color was dominant to green (Y > y) and round seed shape was dominant to shrunken (S > s). (a) What phenotypic ratio would be expected in the F2 from a cross of a pure yellow, round × green, shrunken? (b) What is the F2 ratio of yellow: green and of round: shrunken?

#### Solution 13

1. The ratio of yellow : green = (9/16 yellow, round + 3/16 yellow, shrunken): (3/16 green, round + 1/16 green, shrunken) =12 : 4 = 3 : 1. The ratio of round: shrunken = (9/16 yellow, round + 3/16 green, round) : (9/16 yellow, shrunken + 1/16 green, shrunken) = 12 : 4 = 3 : 1. Thus, at each of the individual loci an F2 phenotypic ratio of 3 : 1 is observed, just as would be expected for a monohybrid cross.

#### Practice 14

How many different crosses may be made (a) from a single pair of factors, (b) from two pairs of factors, and (c) from any given number of pairs of factors (n)?

#### Solution 14

1. All possible matings of the three genotypes produced by a single pair of factors may be represented in a Punnett square.
2. The symmetry of matings above and below the squares on the diagonal becomes obvious. The number of different crosses may be counted as follows: 3 in the first column, 2 in the second, and 1 in the third: 3 + 2 + 1 = 6 different types of matings.

3. There are 32 = 9 different genotypes possible with two pairs of segregating factors. If a 9 × 9 Punnett square were constructed, the same symmetry would exist above and below the squares on the diagonal as was shown in part (a). Again, we may count the different types of matings as an arithmetic progression from 9 to 1; 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45.
4. The sum of any arithmetic progression of this particular type may be found by the formula M = 1/2(g2 + g), where M = number of different types of matings, and g = number of genotypes possible with n pairs of factors.

#### Practice 15

Tall tomato plants are produced by the action of a dominant allele D, and dwarf plants by its recessive allele d. Hairy stems are produced by a dominant gene H, and hairless stems by its recessive allele h. A dihybrid tall, hairy plant is test crossed. The F1 progeny were observed to be 118 tall, hairy : 121 dwarf, hairless : 112 tall, hairless : 109 dwarf, hairy. (a) Diagram this cross. (b) What is the ratio of tall : dwarf; of hairy : hairless? (c) Are these two loci assorting independently of one another?

#### Solution 15

1. Note that the observed numbers approximate a 1 : 1 : 1 : 1 phenotypic ratio.

2. The ratio of tall: dwarf = (118 + 112) : (109 + 121) = 230 : 230 or 1 : 1 ratio. The ratio of hairy: hairless = (118 + 109) : (112 + 121) = 227 : 233 or approximately 1 : 1 ratio. Thus, the testcross results for each locus individually approximate a 1 : 1 phenotypic ratio.
3. Whenever the results of a testcross approximate a 1 : 1 : 1 : 1 ratio, it indicates that the two gene loci are assorting independently of each other in the formation of gametes. That is to say, all four types of gametes have an equal opportunity of being produced through the random orientation that nonhomologous chromosomes assume on the first meiotic metaphase plate. If the testcross does not approximate a 1 : 1 : 1 : 1 ratio, the two genes are probably on the same chromosome (linked). A statistical method for testing genetic hypotheses (e.g., independent assortment) is provided later in this chapter.

#### Practice 16

A dominant allele L governs short hair in guinea pigs and its recessive allele l governs long hair. Codominant alleles at an independently assorting locus specify hair color, such that CYCY = yellow, CYCW = cream, and CWCW = white. From matings between dihybrid short, cream pigs (LlCYCW) predict the phenotypic ratio expected in the progeny.

#### Solution 16

Thus, six phenotypes appear in the offspring in the ratio 3 : 6 : 3 : 1 : 2 : 1. The dash (–) in the genotypes indicates that either allele L or l may be present, with both combinations resulting in a short-haired phenotype.

#### Practice 17

Normal leg size, characteristic of Kerry-type cattle, is produced by the homozygous genotype DD. Shortlegged Dexter-type cattle possess the heterozygous genotype Dd. The homozygous genotype dd is lethal, producing grossly deformed stillbirths called "bulldog calves." The presence of horns in cattle is governed by the recessive allele of another gene locus p, the polled condition (absence of horns) being produced by its dominant allele P. In matings between polled Dexter cattle of genotype DdPp, what phenotypic ratio is expected in the adult progeny?

#### Solution 17

The phenotypic ratio of viable offspring thus becomes: 3 polled : Kerry : 1 horned, Kerry : 6 polled, Dexter : 2 horned, Dexter.

#### Practice 18

(a) A coin is tossed 10 times and lands heads up 6 times and tails up 4 times. Are these results consistent with the expected 50 : 50 ratio? (b) If the coin is tossed 100 times with the same relative magnitude of deviation from the expected ratio, is the hypothesis still acceptable? (c) What conclusion can be drawn from the results of parts (a) and (b)?

#### Solution 18

1. Two mathematical checkpoints are always present in the chi-square calculations: (1) the total of the expected column must equal the total observations, and (2) the sum of the deviations should equal 0. The squaring of negative deviations converts all values to a positive scale. The number of degrees of freedom is the number of phenotypes minus 1 (2 – 1 = 1).We enter Table 2-3 on the first line (df = 1) and find the computed value of 0.4 lying in the body of the table between the values 0.15 and 0.46, corresponding to the probabilities 0.7 and 0.5 shown at the top of the respective columns. This implies that the magnitude of the deviation in our experimental results could be anticipated by chance alone in more than 50% but less than 70% of an infinite number of experiments of comparable size. This range of values is far above the critical probability value of 0.05 or 5%. Therefore, we accept the null hypothesis and conclude that our coin is conforming to the expected probabilities of heads = 1/2 and tails = 1/2.

2. In part (a), heads appeared in 60% and tails in 40% of the tosses. The same relative magnitude of deviations will now be considered in a sample of size of 100. In problems such as this, where expected values are equivalent in all the phenotypic classes, chi-square may be calculated more rapidly by adding the squared deviations and making a single division by the expected number.
3. With df = 1, this X2 value lies between 6.64 and 3.84, corresponding to the probabilities 0.01 and 0.05, respectively. This means that a deviation as large as or larger than the one observed in this experiment is to be anticipated by chance alone in less than 5% of an infinite number of trials of similar size. This is in the ''critical region,'' and we are therefore obliged to reject the null hypothesis and conclude that our coin is not conforming to the expected 50 : 50 ratio. Either of two explanations may be involved: (1) this is not a normal well-balanced coin, or (2) our experiment is among the 1 in 20 (5%) expected to have a large deviation produced by chance alone.

4. The results of parts (a) and (b) demonstrate the fact that large samples provide a more critical test of a hypothesis than small samples. Proportionately larger deviations have a greater probability of occurring by chance in small samples than in large samples.

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