Important Terms

gene (p. 47) allele (p. 47) locus (p. 47) genotype (p. 48) homozygous (p. 48) heterozygous (p. 48) phenotype (p. 48) monohybrid cross (p. 48) P (parental) generation (p. 48) F1 (filial 1) generation (p. 49) reciprocal crosses (p. 49)

F2 (filial 2) generation (p. 49) dominant (p. 51) recessive (p. 51) principle of segregation

(Mendel's first law) (p. 51) concept of dominance (p. 51) chromosome theory of heredity (p. 52) backcross (p. 52) Punnett square (p. 53) probability (p. 54)

multiplication rule (p. 54) addition rule (p. 55) testcross (p. 57) incomplete dominance (p. 58) wild type (p. 58) dihybrid cross (p. 59) principle of independent assortment (Mendel's second law) (p. 60) trihybrid cross (p. 63)

goodness-of-fit chi-square test (p. 66) incomplete penetrance (p. 68) penetrance (p. 68) expressivity (p. 68)

Worked Problems

1. Short hair in rabbits (S) is dominant over long hair (s). The following crosses are carried out, producing the progeny shown. Give all possible genotypes of the parents in each cross.


4 short and 2 long 8 short i2 short

3 short and 1 long 2 long


(a) short X short

(b) short X short

For this problem, it is useful to first gather as much information about the genotypes of the parents as possible on the basis of their phenotypes. We can then look at the types of progeny produced to provide the missing information. Notice that the problem asks for all possible genotypes of the parents.

(a) short X short 4 short and 2 long

Because short hair is dominant over long hair, a rabbit having short hair could be either SS or Ss. The two long-haired offspring must be homozygous (ss) because long hair is recessive and will appear in the phenotype only when both alleles for long hair are present. Because each parent contributes one of the two alleles found in the progeny, each parent must be carrying the s allele and must therefore be Ss.

(b) short X short 8 short

The short-haired parents could be SS or Ss. All 8 of the offspring are short (S_), and so at least one of the parents is likely to be homozygous (SS); if both parents were heterozygous, 1/4 longhaired (ss) progeny would be expected, but we do not observe any long-haired progeny. The other parent could be homozygous (SS) or heterozygous (Ss); as long as one parent is homozygous, all the offspring will be short haired. It is theoretically possible, although unlikely, that both parents are heterozygous (Ss X Ss). If this were the case, we would expect 2 of the 8 progeny to be long haired. Although no long-haired progeny are observed, it is possible that just by chance no long-haired rabbits would be produced among the 8 progeny of the cross.

(c) short X long 12 short

The short-haired parent could be SS or Ss. The long-haired parent must be ss. If the short-haired parent were heterozygous (Ss), half of the offspring would be expected to be long haired, but we don't see any long-haired progeny. Therefore this parent is most likely homozygous (SS). It is theoretically possible, although unlikely, that the parent is heterozygous and just by chance no long-haired progeny were produced.

(d) short X long 3 short and 1 long

On the basis of its phenotype, the short-haired parent could be homozygous (SS) or heterozygous (Ss), but the presence of one long-haired offspring tells us that the short-haired parent must be heterozygous (Ss). The long-haired parent must be homozygous (ss).

Because long hair is recessive, both parents must be homozygous for a long-hair allele (ss).

2. In cats, black coat color is dominant over gray. A female black cat whose mother is gray mates with a gray male. If this female has a litter of six kittens, what is the probability that three will be black and three will be gray?

Because black (G) is dominant over gray (g), a black cat may be homozygous (GG) or heterozygous (gg). The black female in this problem must be heterozygous (Bb) because her mother is gray (gg) and she must inherit one of her mother's alleles. The gray male is homozygous (gg) because gray is recessive. Thus the cross is:

Gg X gg black female gray male

1/2 Ggblack 1/2 gg gray

We can use the binomial expansion to determine the probability of obtaining three black and three gray kittens in a litter of six. Let a equal the probability of a kitten being black and b equal the probability of a kitten being gray. The binomial is (a + b)6, the expansion of which is:

(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6a1 b5 + b6

(See text for an explanation of how to expand the binomial.) The probability of obtaining three black and three gray kittens in a litter of six is provided by the term 20a3b3. The probabilities of a and b are both 1/2, so the overall probability is 20(1/2)3(1/2)3

3. The following genotypes are crossed: AaBbCdDd X AaBbCcDd. Give the proportion of the progeny of this cross having the following genotypes: (a) AaBbCcDd, (b) aabbccdd, (c) AaBbccDd.

This problem is easily worked if the cross is broken down into simple crosses and the multiplication rule is used to find the different combinations of genotypes:

Locus 1 Locus 2

Aa X Aa = / AA, 1/2 Aa, / aa Bb X Bb = /4 BB, /2 Bb, /4 bb

Locus 3 Cc X Cc = 1/4 CC, /2 Cc, 1/4 cc Locus 4 Dd X Dd = /4 DD, / Dd, 1/4 dd

To find the probability of any combination of genotypes, simply multiply the probabilities of the different genotypes:

(a) AaBbCcDd / (Aa) X / (Bb) X / (Cc) X / (Dd) = %6

(b) aabbccdd 1/4 (aa) X /4 (bb) X 1/4 (cc) X 1/4 (dd) = y256

(c) AaBbccDd / (Aa) X / (Bb) X 74 (cc) X / (Dd) = y2

4. In corn, purple kernels are dominant over yellow kernels, and full kernels are dominant over shrunken kernels. A corn plant having purple and full kernels is crossed with a plant having yellow and shrunken kernels, and the following progeny are obtained:

purple, full 112

purple, shrunken 103 yellow, full 91

yellow, shrunken 94

What are the most likely genotypes of the parents and progeny? Test your genetic hypothesis with a chi-square test.

The best way to begin this problem is by breaking the cross down into simple crosses for a single characteristic (seed color or seed shape):

P purple X yellow F1 112 + 103 = 215 purple 91 + 94 = 185 yellow

P purple, full X yellow, shrunken PpFf X ppyy

PpFf = / purple X / full Ppff = 1/2 purple X 1/2 shrunken ppFf = y yellow X 1/2 full ppff = 1/2 yellow X 1/2 shrunken

= 1/4 purple, full = 1/4 purple, shrunken = 1/4 yellow, full = 1/4 yellow shrunken

Our genetic explanation predicts that, from this cross, we should see 1/4 purple, full-kernel progeny; 1/4 purple, shrunken-kernel progeny; 1/4 yellow, full-kernel progeny; and 1/4 yellow, shrunken-kernel progeny. A total of 400 progeny were produced; so 1/4 X 400 = 100 of each phenotype are expected. These observed numbers do not fit the expected numbers exactly. Could the difference between what we observe and what we expect be due to chance? If the probability is high that chance alone is responsible for the difference between observed and expected, we will assume that the progeny have been produced in the 1:1:1:1 ratio predicted by the cross. If the probability that the difference between observed and expected is due to chance is low, the progeny are not really in the predicted ratio and some other, significant factor must be responsible for the deviation.

The observed and expected numbers are:

Phenotype purple full purple shrunken yellow full yellow shrunken


Was this article helpful?

0 0

Post a comment