Supplemental Lecture (98/02/16 update) by Stephen T. Abedon (abedon.1@osu.edu)

  1. Chapter title: Mendelian Genetics Problems
    1. A list of vocabulary words is found toward the end of this document
  2. Vocabulary
    1. No entry.
  3. Practice questions - The below questions are from p. 259-261 of Campbell, 1996:
    1. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white. What is the simplest explanation for the inheritance of these colors in chickens? What offspring would you predict from the mating of a gray rooster and a black hen? [PEEK]
    2. In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: RR (red) x rr (white) --- Rr (pink). If flower position (axial or terminal) is inherited as it is in peas (i.e., axial is dominant to terminal), what will be the ratios of genotypes and phenotypes of the generation resulting from the following cross: axial-red (true-breeding) x terminal-white? What will be the ratios in the F2 generation? [PEEK]
    3. Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:
    4. Character

      Dominant

      Recessive

      Flower position

      Axial (A)

      Terminal (a)

      Stem length

      Tall (L)

      Dwarf (l)

      Seed shape

      Round (R)

      Wrinkled (r)

    5. If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows? (Note: Use the rules of probability instead of a huge Punnett square.) [PEEK]
        1. homozygous for the three dominant traits
        2. homozygous for the three recessive traits
        3. heterozygous
        4. homozygous for axial and tall, heterozygous for seed shape
    6. A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring. [PEEK]
    7. In sesame plants, the one-pod condition (P) is dominant to the three-pod condition (p), and normal leaf (L) is dominant to wrinkled leaf (l). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring: [PEEK]
        1. 318 one-pod normal, 98 one-pod wrinkled
        2. 323 three-pod normal, 106 three-pod wrinkled
        3. 401 one-pod normal
        4. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled
        5. 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod wrinkled
    8. Color patterns in a species of duck is determined by one gene with three alleles. Alleles H and I are semidominant (i.e., incomplete dominance), and allele i is recessive to both. How many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles. [PEEK]
    9. Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following? [PEEK]
        1. all three of their children will be of normal phenotype
        2. one or more of the three children will have the disease
        3. all three children will have the disease
        4. at least one child will be phenotypically normal
    10. The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes? [PEEK]
        1. aabbccdd
        2. AaBbCcDd
        3. AABBCCDD
        4. AaBBccDd
        5. AaBBCCdd
    11. In 1981, a stray black cate with unusual rounded, curled black ears was adopted by a family in California. Hundreds of descendants of the cate have since been born, and cat fanciers hope to develop the "curl" cat into a show breed. Suppose you owned the first culr cat and wanted to develop a true-breeding variety. How would you determine whether the curl allele is dominant or recessive? How would you select for true-breeding cats? How would you know they are true breeding? [PEEK]
    12. What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs)? [PEEK]
        1. AABBCC x aabbcc --- AaBbCc
        2. AABbCc x AaBbCc --- AAbbCC
        3. AaBbCc x AaBbCc --- AaBbCc
        4. aaBbCC x AABbcc --- AaBbCc
    13. In tigers, a recessive allele causes an absence of fur pigmentation (a "white tiger") and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus were mated, what percentatge of their offspring will be cross-eyed? What percentage will be white? [PEEK]
    14. In corn plants, a dominant allele I inhibits kernal color, while the recessive allele i permits color when homozygous. At a different locus, the dominant gene P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the F1 generation? [PEEK]
    15. A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits. Extra digits is a dominant trait. What fraction of this couple's children would be expected to have extra digits? [PEEK]
    16. In mice, black color (B) is dominant to white (b). At a different locus, a dominant allele (A) produces a band of yellow just below the tip of each hair in mice with black fur. This gives a frosted appearance known as agouti. Expression of the recessive allele (a) results in a solid coat color. If mice that are heterozygous at both loci are crossed, what will be the expected phenotypic ratio of their offspring? [PEEK]
    17. The below questions are from p. 621-624 of Keeton, 1976:
    18. If the litter resulting from the mating of two short-tailed cats contains three kittens without tails, two with long tails, and six with short tails, what would be the simplest way of explaining the inheritance of tail length in these cats? Show genotypes (i.e., of all of the individuals mentioned). [PEEK]
    19. In watermelons the genes for green color and for short shape are dominant over their alleles for striped color and for long shape. Suppose a plant with long striped fruit is crossed with a plant heterozygous for both these characters. What phenotypes would this cross produce and in what ratios? [PEEK]
    20. The below questions are from pp. 119-120 of Licata and Garnsey, 1976:
    21. In guinea pigs, black fur (B) is dominant over white fur (b). If one half of a particular litter were white, the genetic makeup of the parents was what? (adapted from G.T. Licata & W.H. Garnsey (1986). A General Review of Biology 2nd Edition. N & N Publishing, Middletown, NY, p. 119 [PEEK]
    22. In sorghum plants, red stem is dominant over green stem. If 1,000 seeds from a sorghum plant germinated to produce 760 red plants and 240 green plants, it would be most reasonable to assume that the parental genotypes were what? (use the abbreviations r and g for red and green alleles, respectively, and irrespective of their dominance relationship) (adapted from G.T. Licata & W.H. Garnsey (1986). A General Review of Biology 2nd Edition. N & N Publishing, Middletown, NY, p. 119 [PEEK]
    23. B is a dominant allele coding for black fur on rabbits and b is a recessive allele coding for white fur on rabbits. Fill in the following blanks with the correct cross of the following: (1) BB x bb, (2) Bb x Bb, (3) bb x bb, (4) Bb x bb (adapted from G.T. Licata & W.H. Garnsey (1986). A General Review of Biology 2nd Edition. N & N Publishing, Middletown, NY, p. 120 [PEEK]
        1. All (100%) of the offspring are white: __________
        2. One quarter (25%) of the offspring are white: __________
        3. All (100%) of the offspring are black: ___________
        4. Three-quarters (75%) of the offspring are black: __________
        5. One-half (50%) of the offspring are white: __________

    24. The below questions are from p. 283 of Lewis, 1998:
    25. A man and a woman each have dark eyes, dark hair, and freckles. The genes for these traits assort independently. The woman is heterozygous for each of these traits, but the man is homozygous. The dominance relationships of the alleles are as follows: [PEEK]
    26. B = dark eyes; b = blue eyes
    27. H = dark hair; h = blond hair
    28. F = freckles; f = no freckles
        1. What is the probability that their child will share the parents' phenotype?
        2. What is the probability that the child will share the same genotype as the mother?
        3. As the father?
    29. (adapted from R. Lewis (1998). Life Third Edition. McGraw-Hill, Boston, Mass., p. 283)
    30. In cats with the Manx trait, the M (dominant) allele causes a short or absent tail, whereas the m allele confers a normal, long tail. Cats of genotype MM die as embryos. If two Manx cats mate, what is the probability that each living kitten has a long tail? (adapted from R. Lewis (1998). Life Third Edition. McGraw-Hill, Boston, Mass., p. 283) [PEEK]
    31. Domesticated hens with white feathers and large, single combs mate with roosters that have dark feathers and small combs. The offspring all resemble their mothers for these two traits. (a) Which alleles are dominant, and which are recessive? (b) If the F1 generation are crossed with each other, what fraction of the F2 generation would be expected to have dark feathers and large, single combs? (adapted from R. Lewis (1998). Life Third Edition. McGraw-Hill, Boston, Mass., p. 283) [PEEK]
    32. The below questions are from p. 304 of Raven and Johnson, 1996:
    33. In 1986, National Geographic magazine conducted a survey of its readers' abilities to detect odors. About 7% of Caucasians in the United States could not smell the odor of musk. If both parents could not smell musk, then none of their children were able to smell it. On the other hand, two parents who could smell musk generally had children who could smell it, too, but a few of the children in those families were unable to smell it. If a single pair of alleles governs this trait, is the ability to smell musk best explained as an example of dominant or recessive inheritance? (adapted from P.H. Raven and G.B. Johnson (1996). Biology. Fourth Edition. Wm. C. Brown Publishers R. Lewis (1998), p. 304) [PEEK]
    34. The below questions are from p. 443 of Gould and Keeton, 1996:
    35. If the dominant allele K is necessary for hearing, and the dominant allele M of another independent loci (i.e., not linked) results in deafness no matter what other genes are present, what percentage of the offspring produced by the cross kkMm x Kkmm will be deaf? (adapted from J.L. Gould and W.T. Keeton (1996). Biological Science. Sixth Edition. W.W. Norton & Company. New York. P. 443) [PEEK]
  4. Practice question answers
    1. P = Gray x Gray; F1 = 15 Gray, 6 black, 8 white. Say that BX = black, WX = white, and WB = gray. Then a WB x WB should result in a ratio of 1:2:1 for white (WW), gray (WB), and black (BB), respectively. The ratio observed is 1.1:2.0:0.8. Therefore I would conclude that the simplest explanation is the semidominance (i.e., incomplete dominance) where gray is the heterozygote (one loci, two alleles). The phenotypes of the offspring which would arise from a matting between a gray rooster and a black hen would have to been half black and half gray since we know that the black hen has to be a homozygote (otherwise she would be grey).
    2. The genotype of a true breeding axial-red must be AA RR (where A is the allele which codes for axial and R is the allele which codes for red). Similarly, the terminal-white must be TT WW (where T and W code for terminal and white, respectively) because the T allele is recessive, and the W allele results in white only in the homozygote. A cross between these two will have only one genotype (AT RW) and one phenotype (axial-pink). Note that these are dihybrids. The F2 generation, however, will diverge from the expected phenotypic ratio of 9:3:3:1 due to a lack of pure dominant-recessive relationship (i.e., one loci demonstrates semidominance, a.k.a., incomplete dominance). To determine the various expected genotype and phenotype ratios, start with a knowledge that there will be an equiprevalence of the following gametes: AR, AW, TR, and TW. Then simply determine genotype ratios employing a Punnett square, finally deriving phenotype ratios from your derivation of genotype ratios.
    3. We know that the plant is heterozygous for all three loci. Therefore we know that each plant has six unique alleles. We'll assume a lack of linkage between loci. We'll also, therefore, assume independent assortment. (i) A homozygote would be the product of identical gametes. Note that there are 23 possible homozygotes (i.e., 8) since there are three loci and two possible alleles at each loci. There is a relatively simple way of figuring out the answer to this question. First determine how many possible gametes may be produced. The answer, once again, is eight (for the same reason as above). If you started with one gamete of any kind, what would be the odds of picking a second gamete of the same type? We're assuming that the gamete pool is note depleted, so the odds of removing a second gamete of the same type are one in eight, which is the answer to this question. (ii) This is a different question, and a relatively easy one to answer. Now instead of eight possible homozygote combinations, there is only one possible. The odds of picking an all-recessive gamete in the first round are one in eight. The odds of picking an all-recessive gamete in the second round are also one in eight. Consequently, the odds of picking both are one in sixty four (8 x 8), which is the answer to this question. (iii) I'll assume that what they are looking for here is the trihybrid. The odds of picking the trihybrid would be the same as those for picking the all recessive, i.e., one in 64. The odds of picking an individual which is a heterozygote at at least one loci is simple 1 minus the odds of picking an all homozygote. Thus, of 64 progeny, 64 - 8 would be a heterozygote to at least some degree, or seven in eight. (iv) This is one is relatively easy to answer. It is simply 2 in 64, or 1 in 32. This is because the odds of picking the gamete which is ALR is one in eight and the odds of then picking the gamete which is ALr is also one in eight, making the odds of picking both, in this order one in 64. However, these gametes could be picked also in the reverse order, which makes the odds of coming up with this zygote two times one in 64, or 1 in 32.
    4. The first black guinea pig was homozygous for fur color, the second was a heterozygote. Called the black allele B and the albino allele O, the parental genotypes are BB, OO, and BO, respectively. The first guinea pig produced only B gametes, the second only O gametes, and the third both gamete types in a 1:1 ratio. The offspring from the first mating all had the genotype BO. The offspring from the second mating were nearly 1:1 BO and OO.
    5. (i) PPLl x PPLl. (ii) ppLl x ppLl. (iii) PPLL x whatever or PP x LL (not worrying about the not shown loci). (iv) To find the answer to this one, first look at the phenotypes associated with each loci individually. You will see that there are 3 one-pods for each three-pod and 1 normal for each wrinkled. Thus, you might start with the assumption that you have a cross between two heterozygotes at the first loci and a heterozygote and a homozygote (recessive) at the second loci. Thus, the cross has to be Ppll x PpLl. (v) This is approximately a 9:3:1:1 ratio, which is the phenotypic ratio following a dihybrid cross (with the dominant phenotype present in largest numbers and the dual recessive least prevalent), i.e., PpLl x PpLl.
    6. Possible genotypes are HH, HI, Hi, II, Ii, and ii. Note that Hi has the same phenotype as HH and Ii has the same phenotype as II. Therefore there are six minus two possible phenotypes. Note that the genetics of this system are identical to the genetics of the ABO blood group.
    7. For a carrier the probability of producing a carrying gamete is 0.5 and the probability of producing a not carrying gamete is also 0.5. (i) This is equal to the odds of not producing an affected child, cubed. The odds of not getting an affected child is simply a Mendelian ratio result, i.e., the dominant phenotype (no PKU) will be present on average 75% of the time. Thus, the answer is 0.753 (which is 0.42). (ii) To answer this you simply have to figure out what the odds of having three not afflicted children (i.e., 0.42) and subtracting this number from 1 (i.e., the answer is 0.58). (iii) This is equal to the odds of getting the disease, cubed. That is, (0.25)3 or 0.016. (iv) This simply 1 - (0.25)3 (i.e., 0.98).
    8. To answer these first note that the frequency of the least prevalent progeny genotype is 1 in 28 (1/256). Why? There are two possible alleles per loci, four loci, and two alleles per individual. In addition, there are 24 (16) types of gametes. (i) This all lower case individual can be produced only by the fusion of two abcd gametes. The odds of this occurring are equal to 1 in (24 x 24) = 1 in 28 = 1 in 256. That is, this genotype should be minimally prevalent. (ii) This tetrahybrid individual may be produced by the fusion of a variety of gamete combinations including (assume that order is important, e.g., male gametes come first and female gametes second): ABCD x abcd, abcd x ABCD, AbCD x aBcd, aBcd x AbCD, etc. The only consistent theme is that for each loci, the first gamete must have a different allele than is carried by the second gamete. For each loci the odds of this are one-half. Thus, the odds of picking a second gamete which differs from the first at each loci are (0.5)4 (4 for 4 loci) = 0.0625. Since you can start with any gamete type (i.e., odds equal 1 in 1), this 0.0625 (or 1 in 16) is the odds of producing a tetrahybrid from a tetrahybrid cross. (iii) This is the same as the all lower case, i.e., 1 in 256. (iv) Note that this is heterozygous at two loci, but homozygous at the other two. Thus, the first gamete is constrained by having both a B and a c. The odds of producing this gamete are therefore 0.5 x 0.5 = 0.25, so one-quarter of first gametes chosen could produce this diploid (i.e., this contrasts with the second example, for which all gametes could have served as the first gamete). The second gamete is 100% constrained by the first gamete, so its odds of being chosen are 1 in 16 ((0.5)4) = 0.0625. The odds of producing this individual therefore are 0.25 x 0.0625 = 0.016 = 1/64. That is, four of 256 progeny will have this genotype. (v) Using the same reasoning as in the previous example, the odds of producing the first gamete are (0.5)3 = 1 in 8 = 0.125. Again, the second gamete is 100% constrained so the odds of producing this gamete are 0.125 x 0.0625 = 1 in 128, or 2 of 256 progeny.
    9. The name of the game here is backcrossing and other forms of inbreeding, plus diagnostic test crossing. First, you don't know whether the allele is dominant or recessive. However, you do know a priori that it will be easier to develop a true breeding recessive than it will be to develop a true breeding dominant character, if for no other reason than that any individual displaying a recessive character must be homozygous (i.e., true breeding). Regardless, the first thing you are going to have to do is to cross the curl cat with a second, preferably unrelated cat. If the curl is carried by a recessive allele and the allele is not carried by the non-curled cat, then all progeny from this test mating will lack the curl, but all will be heterzygous at the curl loci. If the curl is recessive and a curl allele is carried by the non-curl cat, then half the progeny will have the curl and half will not. However, if the curl is dominant, then chances are that half the progeny will also carry the curl and half will not (since the non-curl cat would definitely lack the curl, but the curl cat likely is a heterozygote since the allele presumably is rare). If the curl is recessive and the non-curl cat is a heterozygote, then the non-curl progeny also must all be heterozygotes, but homozygotes if the curl is dominant. Distinguishing these possibilities will require a second test cross e.g., among non-curl progeny. As noted, selection for true breeding cats would require back crossing and other forms of inbreeding. You would determine whether individuals are true-breeding based on the results of test crosses like those described above.
    10. (i) 1.0 (= 1.0 x 1.0 x 1.0), (ii) 0.5 x 0.25 x 0.25 = 0.031, (iii) 0.5 x 0.5 x 0.5 = 0.125, (iv) 1.0 x 0.5 x 1.0 = 0.5.
    11. If the same allele causes both conditions, i.e., white and cross-eyed, then the two phenotypes should completely overlap. Thus, a mating between two heterozygotes should show the normal Mendelian ratio for both phenotypes, and thus 25% of progeny will be cross-eyed, and the same 25% will be white.
    12. Heterozygous at each loci means Ii Pp. This is simply a cross between dihybrids, except that the various alleles do not have simple dominant-recessive relations, and their exists an interaction between the two loci. To solve this problem you are well off using a Punnett square to determine progeny genotypes, and then simply figure out what phenotype is associated with each genotype.
    13. IP
      0.25
      Ip
      0.25
      iP
      0.25
      ip
      0.25
      IP
      0.25
      IIPP
      0.0625
      white
      IIPp
      0.0625
      white
      IiPP
      0.0625
      white
      IiPp
      0.0625
      white
      Ip
      0.25
      IIPp
      0.0625
      white
      IIpp
      0.0625
      white
      IiPp
      0.0625
      white
      Iipp
      0.0625
      white
      iP
      0.25
      IiPP
      0.0625
      white
      IiPp
      0.0625
      white
      iiPP
      0.0625
      purple
      iiPp
      0.0625
      purple
      ip
      0.25
      IiPp
      0.0625
      white
      Iipp
      0.0625
      white
      iiPp
      0.0625
      purple
      iipp
      0.0625
      red

      So the ratio of white:purple:red kernels is 12:3:1.

    14. One-half. Why? The wife and daughter we know are homozygous recessive from their phenotype. If this man can father a child which does not have the dominant allele, then the man must be a heterozygote. Therefore, half of his children will carry the dominant allele, so half will express the dominant phenotype.
    15. Bb Aa x Bb Aa is the cross. The possible phenotypes are black (BX aa), white (bb XX), and agouti (BX AX). The odds of getting a white mouse therefore is 1 in 4 or 25% (i.e., independent of the A locus, the B locus will be homozygous recessive 25% of the time). This of course means that the mice will be either black or agouti 75% of the time. Note that to be black the A locus must also be homozygous recessive. Thus, the probability that a mouse will be black is 0.25 x 0.75 = 0.1875. The probability that a mouse will be agouti, therefore, must be 1.0 - 0.1875 - 0.25 = 0.5625.
    16. You would argue that short tails is the trait exhibited by the heterozygote, and long and short tails by the two homozygotes. That is, you would argue that inheritance of tail length in cats is controlled by a one loci, two allele system which displays semidominance (i.e., incomplete dominance). If a double dose of the allele T results in long tails and a double dose of the allele N results in no tails, then short tails would be associated with the genotype TN.
    17. The striped, long fruit is clearly homozygous recessive for both traits. Call the alleles G, g, s, S for green, striped, long, and short, respectively. Thus, the just described parent has the genotype gg ss and the heterozygous parent (the dihybrid) has the genotype Gg Ss. In the F1 generation there would be a total of four genotypes, each equi-present. These would be Gg Ss, gg Ss, Gg ss, and gg ss corresponding to the phenotypes: green and short, striped and short, green and long, and striped and long.
    18. Bb x bb.
    19. rg x rg
    20. (i) 3, (ii) 2, (iii) 1, (iv) 2, (v) 4.
    21. (i) 1.0 (100%), (ii) 1/2 = 1/8, (iii) same as (ii)
    22. 1/3
    23. (i) the allele for white feathers is dominant to the allele for dark feathers and the allele for large, single combs is dominant to the allele for small combs; (ii) this is a dihybrid cross and the question is what fraction of the F2 generation possess one dominant and one recessive trait---the answer is 3/16 (recalling, of course, the Mendelian ratio, 9:3:3:1)
    24. This is probably an example of recessive inheritance.
    25. half will be Kk and half will be mm. How many will be Kkmm and therefore able to hear? 0.5 * 0.5 = 0.25, one-quarter. Therefore, three-quarters will be deaf.
  5. References
    1. Campbell, N. A. (1996). Biology. Fourth Edition. Benjamin/Cummings Publishing, Menlo Park, California. pp. 259-261.
    2. Keeton, W. T. (1976). Biological Science. W. W. Norton & Co., New York. pp. 621-624.