WORKED PROBLEMS

Problem 1

If a fertilized Drosophila egg is punctured at the anterior end and a small amount of cytoplasm is allowed to leak out, what will be the most likely effect on the development of the fly embryo?

Solution Strategy

What information is required in your answer to the problem?

Likely effects on development of removing cytoplasm from the anterior end of a fertilized fly egg.

What information is provided to solve the problem?

Cytoplasm is removed from the anterior end.

For help with this problem, review:

Egg-polarity genes in Section 22.2.

Solution Steps

The egg-polarity genes determine the major axes of development in the Drosophila embryo. One of these genes is bicoid, which is transcribed in the maternal parent. As bicoid mRNA passes into the egg, the mRNA becomes anchored to the anterior end of the egg. After the egg has been laid, bicoid mRNA is translated into Bicoid protein, which forms a concentration gradient along the anterior–posterior axis of the embryo. The high concentration of Bicoid protein at the anterior end induces the development of anterior structures such as the head of the fruit fly. If the anterior end of the egg is punctured, cytoplasm containing high concentrations of Bicoid protein will leak out, reducing the concentration of Bicoid protein at the anterior end. The result will be that the embryo fails to develop head and thoracic structures at the anterior end.

Problem 2

The immunoglobulin molecules of a particular mammalian species have kappa and lambda light chains and heavy chains. The kappa gene consists of 250 V and 8 J segments. The lambda gene contains 200 V and 4 J segments. The gene for the heavy chain consists of 300 V, 8 J, and 4 D segments. If just somatic recombination and random combinations of light and heavy chains are taken into consideration, how many different types of antibodies can be produced by this species?

Solution Strategy

What information is required in your answer to the problem?

The number of different types of antibodies that can be produced if somatic recombination and random combinations of light and heavy chains are considered.

What information is provided to solve the problem?

• The kappa gene has 250 V and 8 J segments.
• The lambda gene has 200 V and 4 J segments.
• The heavy chain has 300 V, 8 J, and 4 D segments.

For help with this problem, review:

The Generation of Antibody Diversity in Section 22.6.

Solution Steps

For the kappa light chain, there are 250 × 8 = 2000 combinations; for the lambda light chain, there are 200 × 4 = 800 combinations; so a total of 2800 different types of light chains are possible. For the heavy chains, there are 300 × 8 × 4 = 9600 possible types. Any of the 2800 light chains can combine with any of the 9600 heavy chains; so there are 2800 × 9600 = 26,880,000 different types of antibodies possible from somatic recombination and random combination alone. Junctional diversity and somatic hypermutation would greatly increase this diversity.