Patterns of Inheritance Practice Problems

Practice 4

Coat colors of the Shorthorn breed of cattle represent a classical example of codominant alleles. Red is governed by the genotype C^RC^R, roan (mixture of red and white) by C^RC_W, and white by C^WC^W. (a) When roan Shorthorns are crossed among themselves, what genotypic and phenotypic ratios are expected among their progeny? (b) If red Shorthorns are crossed with roans, and the F¹ progeny are crossed among themselves to produce the F², what percentage of the F² will probably be roan?

Solution 4

1. 

Since each genotype produces a unique phenotype, the phenotypic ratio 1 : 2 : 1 corresponds to the same genotypic ratio.

2. 

There are three types of matings possible for the production of the F₂. Their relative frequencies of occurrence may be calculated by preparing a mating table.



1. The mating C^RC^R × C^RC^R (red × red) produces only red (C^RC^R) progeny. But only one-quarter of all matings are of this type. Therefore, four of all the F2 should be red from this source.
2. The matings C^RC^W × C^RC^R (roan female × red male or roan male × red female) are expected to produce 1/2 C^RC^R (red) and 1/2 C^RC^w (roan) progeny. Half of all matings are of this kind. Therefore, (1/2) (1/2) 1/4 of all the F² progeny should be red and four should be roan from this source.
3. The mating C^RC^W × C^RC^W (roan × roan) is expected to produce 1/4 C^RC^R (red), 1/2 C^RC^W (roan), and 1/4 C^WC^W (white) progeny. This mating type constitutes 1/4 of all crosses. Therefore, the fraction of all F¹ progeny contributed from this source is (1/4) (1/4) = 1/16 C^RC^R, (1/4) (1/2) = 1/8 C^RC^W , (1/4) (1/4) = 1/16 C^WC^W.

The expected F² contributions from all three types of matings are summarized in the following table.

The fraction of roan progeny in the F² is 3/8, or approximately 38%.

Practice 5

The absence of legs in cattle ("amputated") has been attributed to a completely recessive lethal gene. A normal bull is mated with a normal cow and they produce an amputated calf (usually dead at birth). The same parents are mated again. (a) What is the chance of the next calf being amputated? (b) What is the chance of these parents having two calves, both of which are amputated? (c) Bulls carrying the amputated allele (heterozygous) are mated to noncarrier cows. The F¹ is allowed to mate at random to produce the F². What genotypic ratio is expected in the adult F²? (d) Suppose that each F¹ female in part (c) rears one viable calf, i.e., each of the cows that throws an amputated calf is allowed to remate to a carrier sire until she produces a viable offspring. What genotypic ratio is expected in the adult F²

Solution 5

1. If phenotypically normal parents produce an amputated calf, they must both be genetically heterozygous.



Thus, there is a 25% chance of the next offspring being amputated.

2. The chance of the first calf being amputated and the second calf also being amputated is the product of the separate probabilities: (1/4) (1/4) = 1/16 (see Probability Theory later in this chapter).
3. The solution to part (c) is analogous to that of Problem 2.4(b). A summary of the expected F₂ is as follows:



All aa genotypes die and fail to appear in the adult progeny. Therefore, the adult progeny has the genotypic ratio 9AA : 6Aa or 3AA : 2Aa.

4. The results of matings AA × AA and AA × Aa remain the same as in part (c). The mating of Aa × Aa now is expected to produce 1/3AA and 2/3Aa adult progeny. Correcting for the frequency of occurrence of this mating, we have (1/4) (1/3) = 1/12AA and (1/4) (2/3) = 2/12Aa.
   Summary of the F²:



The adult F² genotypic ratio is expected to be 7AA: 5Aa.

Practice 6

The genetics of rabbit coat colors is given in Example 2.20. Determine the genotypic and phenotypic ratios expected from mating full-colored males of genotype Cc^ch to light-gray females of genotype c^chc.

Solution 6

Thus, we have a 1 : 1 : 1 : 1 genotypic ratio, but a phenotypic ratio of 2 full color : 1 chinchilla : 1 light gray.

Practice 7

A man is suing his wife for divorce on the grounds of infidelity. Their first child and second child, whom they both claim, are blood groups O and AB, respectively. The third child, whom the man disclaims, is blood type B. (a)Can this information be used to support the man's case? (b) Another test was made in the M-N blood group system. The third child was group M, the man was group N. Can this information be used to support the man's case?

Solution 7

1. The genetics of the ABO blood group system was presented in Example 2.19. Because the group O baby has the genotype ii, each of the parents must have been carrying the recessive allele. The AB baby indicates that one of the parents had the dominant I^A allele and the other had the codominant allele I^B. Any of the four blood groups can appear among the children whose parents are I^Ai × I^Bi. The information given on ABO blood groups is of no use in supporting the man's claim.
2. The genetics of the M-N blood group system was presented in Example 2.12. The M-N blood groups are governed by a pair of codominant alleles, where groups M and N are produced by homozygous genotypes. A group N father must pass the L allele to his offspring; they all would have the N antigen on their red blood cells, and would all be classified serologically as either group MN or N depending upon the genotype of the mother. This man could not be the father of a group M child.

Practice 8

Heterozygous black guinea pigs (Bb) are crossed among themselves. (a)What is the probability of the first three offspring being alternately black-white-black or white-black-white? (b) What is the probability among three offspring of producing two black and one white in any order?

Solution 8

1. 
2. Consider the number of ways that two black and one white offspring could be produced.

Once we have ascertained that there are three ways to obtain two black and one white, the total probability becomes 3(3/4)² (1/4) = 27/64.

Practice 9

Expand the binomial (p + q)⁵.

Solution 9

1st term: 1p⁵q⁰; coefficient of 2nd term = (5)(1)/1
2nd term: 5p⁴q¹; coefficient of 3rd term = (4)(4)/2
3rd term: 10p³q²; coefficient of 4th term = (3)(10)/3
4th term: 10p²q³; coefficient of 5th term = (2)(10)/4
5th term: 5p¹q⁴; coefficient of 6th term = (1)(5)/5
6th term: 1p⁰q⁵

Summary: (p + q)⁵ = p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵

Practice 10

Find the middle term of the expansion (p + q)¹⁰ by application of formula (2.2).

Solution 10

The middle term of the expansion (p + q)¹⁰ is the sixth term since there are (n + 1) terms in the expansion. The power of q starts at zero in the first term and increases by 1 in each successive term so that the sixth term would have q⁵, and so k = 5. Then the sixth term is

Practice 11

A multiple allelic series is known with seven alleles. How many kinds of matings are possible?

Solution 11

Practice 12

The MN blood types of humans are under the genetic control of a pair of codominant alleles as explained in Example 2.12. In families of six children where both parents are blood type MN, what is the chance of finding three children of type M, two of type MN, and one of type N? [Hint: use formula (2.4).]

Solution 12

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. 
2. 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.



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.

2. 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.
3. 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 C^YC^Y = yellow, C^YC^W = cream, and C^WC^W = white. From matings between dihybrid short, cream pigs (LlC^YC^W) 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.



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.

3. 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.

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 (o – e) 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 |o – e| 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.