Patterns of Inheritance Practice Problems (page 2)

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Updated on Aug 19, 2011

Practice 4

Coat colors of the Shorthorn breed of cattle represent a classical example of codominant alleles. Red is governed by the genotype CRCR, roan (mixture of red and white) by CRCW, and white by CWCW. (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 CRCR × CRCR (red × red) produces only red (CRCR) 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 CRCW × CRCR (roan female × red male or roan male × red female) are expected to produce 1/2 CRCR (red) and 1/2 CRCw (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 CRCW × CRCW (roan × roan) is expected to produce 1/4 CRCR (red), 1/2 CRCW (roan), and 1/4 CWCW (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 CRCR, (1/4) (1/2) = 1/8 CRCW , (1/4) (1/4) = 1/16 CWCW.

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

  3. 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).
  4. The solution to part (c) is analogous to that of Problem 2.4(b). A summary of the expected F2 is as follows:
  5. 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.

  6. 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:
  7. 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 Ccch to light-gray females of genotype cchc.

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 IA allele and the other had the codominant allele IB. Any of the four blood groups can appear among the children whose parents are IAi × IBi. 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. 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).]

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