Section 5.7: Multi-trait Inheritance
Tutorial 1: Solving Dihybrid Cross Problems
Practice
(Page 212)
1. (a) The possible gametes for parent 1 are shown in the diagram below:
The possible gametes for parent 2 are shown in the diagram below:
Each parent can produce only two types of gametes, so the Punnett square needs only a 2 X 2 grid:
The possible phenotypes are:
Barking, erect: BbEe (total = 1)
Barking, drooping: Bbee (total = 1)
Non-barking, erect: bbEe (total = 1)
Therefore, the phenotypic ratio of the offspring is 1:1:1:1.
(b) The possible gametes for parent 1 are shown in the diagram below:
The possible gametes for parent 2 are shown in the diagram below:
Copyright © 2011 Nelson Education Ltd.
Chapter 5: Mendelian Genetics—Patterns of Inheritance
5.7-1
Parent 1 can produce only one type of gametes and parent 2 can produce only two types, so the Punnett
square needs only a 2 X 1 grid:
The possible phenotypes are:
Barking, drooping: Bbee (total = 1)
Non-barking, drooping: bbee (total =1)
Therefore, the phenotypic ratio of the offspring is 1:1.
(c) Both parents have the same genotype, and so the possible gametes for each parent are:
Since both parents can produce four types of gametes, the Punnett square must have a 4 X 4 grid:
The possible phenotypes are:
Barking, erect: BBEE, BBEe, BbEE, BbEe, BBEe, BbEe, BbEE, BbEe, BbEe (total = 9)
Barking, drooping: BBee, Bbee, Bbee (total = 3)
Non-barking, erect: bbEE, bbEe, bbEe (total = 3)
Non-barking, drooping: bbee (total =1)
Therefore, the phenotypic ratio of the offspring is 9:3:3:1.
(d) The possible gametes for parent 1 are:
Copyright © 2011 Nelson Education Ltd.
Chapter 5: Mendelian Genetics—Patterns of Inheritance
5.7-2
The possible gametes for parent 2 are:
Parent 1 can produce only two types of gametes and parent 2 can produce only one type, so the Punnett
square needs a 4 X 1 grid:
The possible phenotypes are:
Barking, erect: BbEe (total = 1)
Barking, drooping: Bbee, (total = 1)
Non-barking, erect: bbEe (total = 1)
Non-barking, drooping: bbee (total = 1)
Therefore, the phenotypic ratio of the offspring is 1:1:1:1.
Section 5.7 Questions
(Page 214)
1. Mendel’s law of independent assortment states that if genes are located on separate chromosomes, they
will be inherited independently of each other. Examples may vary, but will likely be drawn from the
section, such as the dihybrid cross of round, yellow peas with wrinkled, green peas.
2. (a) The possible gametes for this individual are We and we.
(b) The possible gametes for this individual are Ea and ea.
(c) The possible gametes for this individual are tY.
3. (a) The genotype of the mother is CcAa and the genotype of the father is CCaa.
The possible gametes for the mother are shown in the diagram below:
The possible gametes for the father are shown in the diagram below:
The mother can produce four types of gametes and the father can produce only one type, so the Punnett
square needs a 1 × 4 grid:
Copyright © 2011 Nelson Education Ltd.
Chapter 5: Mendelian Genetics—Patterns of Inheritance
5.7-3
4.
5.
6.
7.
The possible phenotypes are:
Curly, normal pigmentation: CCPp, CcPp (total = 2)
Curly, albinism: CCpp, Ccpp (total = 2)
Their child will be either curly-haired with albinism or curly-haired with normal pigmented skin.
(b) From the Punnett square, the probability of each phenotype is:
Curly, normal pigmentation: CCPp, CcPp, CcPp (2/4 = ½)
Curly, albinism: CCpp (2/4 = ½)
The probability of the child being curly-haired with albinism is ½ (50 %). The probability of the child
being curly-haired with normal pigmented skin is ½ (50 %).
(c) The sex of the child is determined by the presence of a Y chromosome and is independent of hair
curliness or skin pigmentation. Only the father can donate a Y chromosome, so there is a ½ chance that the
child is a boy.
From (b), we know that the probability that a child will be curly-haired with albinism is ½. Using the
product law, the probability that the child will be a male with curly-hair and albinism is: ½ × ½ = ¼
(25 %).
(a) Since B is the dominant allele, both parents may be either homozygous (BB) or heterozygous (Bb) and
have black offspring (B_). However, since all the offspring are black, it is probable that at least one parent
is homozygous (BB). Since T is the dominant allele and all the offspring have the recessive pacing trait
(tt), the parents can be either heterozygous (Tt) or homozygous (tt). Since all the offspring are pacing, it is
more likely that both parents are homozygous (tt).
(b) B is the dominant allele (for black) and T is a dominant allele (for trotting). Colt 1 is black and
trotting, so it must be either Bbtt or Bbtt. The stud and mare 1are both black, so they both may be
heterozygous or homozygous for B (BB or Bb). Both the stud and mare 1 donated a t allele to colt 1. Since
they are both trotting, the stud and mare 1 must be heterozygous for this gene (Tt). The genotype of mare 1
must be either BbTt or BBTt. According to this cross, the genotype of the stud is B_Tt.
Mare 2 is chestnut and pacing, both of which are recessive traits. Her genotype is therefore bbtt. Colt 2 is
chestnut and trotting. Colt 2 could only get the dominant allele T, from the stud. Therefore, the genotype
of colt 2 is bbTt. Since colt 2 is chestnut, the stud must have donated one b allele. Therefore, the genotype
of the stud can only be BbTt.
Therefore, the genotypes are:
stud, BbTt; mare 1, BbTt or BBTt; colt 1, BBtt or Bbtt; mare 2, bbtt; and colt 2, bbTt
Both continuous and discontinuous variation are terms that are used to describe the interaction between
two or more genes. In continuous variation, the product of one gene is affected by the product(s) of one or
more other genes. In discontinuous variation, there is no affect of the product of one gene by the product
of another gene.
Answers will vary, but students should outline an experiment in which a variable characteristic of tomato
plants is followed in a defined cross between true-breeding plants to observe if the phenotypes of the
offspring show continuous or discontinuous variation.
Sample answer: This investigation will determine if the characteristic of fruit shape (round versus oval)
shows continuous or discontinuous variation. Cross true-breeding round-fruited tomato plants with truebreeding oval-fruited plants. These are the parental (P) generation. Plant out 100 or more of the seeds from
the cross and observe their characteristics. This is the F1 generation. If the traits show discontinuous
variation, the F1 generation will all have either oval or round fruit. If the traits show continuous variation,
the fruit of the F1 generation will show a variation in shapes between round and oval.
The genotype of the man is IAIARh+Rh. Since blood type O (and Rh) is recessive, the woman’s
genotype is iiRh Rh. The possible gametes from these individuals are:
Copyright © 2011 Nelson Education Ltd.
Chapter 5: Mendelian Genetics—Patterns of Inheritance
5.7-4
Therefore, the possible genotypes of their child is
The possible phenotypes for their child are blood type A, Rh or blood type B, Rh+.
8. Sample size is important because it is easier to see patterns and trends with larger numbers. With larger
sample sizes, random variation has a smaller effect on the results. The researcher can then be more
confidence that the observed pattern is a result of the cross, and not random chance.
9. Yes, it is possible to have a trihybrid cross. Students explanations may vary.
Sample answer: A trihybrid cross would require the production of parents that were heterozygous for
three traits that are determined by genes on different chromosomes. This would be more difficult than a
dihybrid cross, but not impossible. The parents would produce eight different types of gametes. The
expected genotypes and phenotypes could then be determined using an 8 × 8 Punnett square.
Copyright © 2011 Nelson Education Ltd.
Chapter 5: Mendelian Genetics—Patterns of Inheritance
5.7-5