13 Jun Suppose an individual is heterozygous for both traits
Question
14. Suppose an individual is heterozygous for both traits (eye pigmentation and chin form).
a. What is the genotype of such an individual?
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b. What are the possible genotypes of that individual’s gametes?
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If determining the answer for the last question was difficult, recall that the principle of independent assortment states that genes on different chromosomes are separated out independently of one another during meiosis. That is, the occurrence of an allele for eye pigmentation in a gamete has no bearing on which allele for chin form will occur in that same gamete.
There is a useful convention for determining possible gamete genotypes produced during meiosis from a given parental genotype. Using the genotype PpDd as an example,
here’s the method:
Using PpDd, match the first allele (P) to the third allele (D) = PD
match the first allele (P) to the fourth allele (d) = Pd
match the second allele (p) to the third allele (D) = pD
match the second allele (p) to the fourth allele (d) = pd
therefore use: 1-3 (PD)
1-4 (Pd)
2-3 (pD)
2-4 (pd)
c. Suppose two individuals heterozygous for both eye pigmentation and chin form have children. What are the possible genotypes of their children?
You can set up a Punnett square to do dihybrid problems just as you did with monohybrid problems. However, depending upon the parental genotypes, the square may have as many as 16 boxes rather than just 4. Insert the possible genotypes of the gametes from one parent in the top circles and the gamete genotypes of the other parent in the circles to the left of the box.
Gametes of First Parent
Gametes of the Second Parent
d. What is the genotypic ratio?
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e. What is the phenotypic ratio?
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15. You would probably agree that it is unlikely that a family will have 16 children. In fact, one of the most useful facets of problems such as these is that they allow you to predict what the chances are for a phenotype occurring. Genetics is really a matter of probability, the likelihood of the occurrence of any particular outcome.
To take a simple example, consider that the probability of coming up with heads in a single toss of a coin is one chance in two, or 1/2.
Now apply this example to the question of the probability of having a certain genotype. Look at your Punnett square in problem 15. The probability of having a genotype is the sum of all occurrences of that genotype. For example, the genotype PPDD occurs in 1 of the 16 boxes. The probability of having the genotype PPDD is 1/16.
f. What is the probability of an individual from the above problem having the genotype:
ppDD ____________________________________________________
PpDd ____________________________________________________
PPDd ____________________________________________________
To extend this idea, let’s consider the probability of flipping heads twice in a row with our coin. The chance of flipping heads the first time is 1/2. The same is true for the second flip. The chance (probability) that we will flip heads twice in a row is 1/2 X 1/2 = 1/4. The probability that we could flip heads three times in a row is 1/2 X l/2 X 1/2 = 1/8.
g. Returning to eye color and chin form, state the probability that three children born to these parents will have the genotype ppdd.
h. What is the probability that three children born to these parents will have pigmented eyes and dimpled chins?
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i. What is the genotype of the F1 generation when the father is homozygous for both pigmented eyes and dimpled chin, and the mother has blue eyes and no dimple?
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j. What is the phenotype of the individual(s) you determined in letter i, above?
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