Biological
Sciences Answers & Explanations
1. D
Bacteria are often classified and named on the basis of their
shape, of which there are three: spherical, rod-like, and helical
(spiral). Spherical bacteria are known as cocci; rod-like
bacteria are known as bacilli; and helical bacteria are known as
spirochetes, or spirilla. Thus, if a sample from a CF patient
infected with Staphylococcus were cultured and viewed
under a microscope, the bacteria would appear spherical in
shape.
Helical bacteria are the least common of the three groups;
rod-shaped bacteria include the common E.coli, as well as those
bacteria responsible for causing lock-jaw, diphtheria, and
tuberculosis. So, Choice A and Choice C are wrong. Choice B is
wrong since sickle-shaped bacteria do not exist; however, those
individuals with the genetic disease sickle-cell anemia have red
blood cells with an abnormal sickle shape.
2. C
CF patients suffer from pancreatic insufficiency because of
abnormally viscous mucus secretions that block the ducts linking
the pancreas to the small intestine. The pancreas synthesizes and
secretes the following enzymes: trypsinogen, chymotrypsinogen,
carboxypeptidase, amylase, and lipase.
Lipase digests fats, amylase digests carbohydrates;
carboxypeptidase, trypsinogen, and chymotrypsinogen all digest
proteins. Trypsinogen and chymotrypsinogen are secreted into the
small intestine, where enterokinase, an enzyme secreted by the
glands of the small intestine, converts trypsinogen to its active
form, trypsin. Trypsin then converts chymotrypsinogen to its
active form, chymotrypsin. Choice C is therefore correct, since
enterokinase - a product of the small intestine--would not be
lacking in a CF patient. On the other hand, lipase, trypsin, and
chymotrypsinogen would be included in an enzyme supplement
administered to CF patients. Thus, Choice A, Choice B, and Choice
D are all incorrect.
3. B
According to the passage, the gene for CF is autosomal
recessive. Let F = the normal gene and f = the CF gene. Those
individuals with CF have the genotype ff; those who are carriers
have the genotype Ff; and normal individuals have the genotype
FF. So in a cross between the two heterozygotes: Ff x Ff, 25
percent of the offspring will have the genotype FF and will be
normal; 50 percent will have the genotype Ff and will thus be
carriers; and 25 percent will have the genotype ff and will have
CF. Therefore, in a cross between two carriers, there is a 25
percent chance that their child will be affected by CF.
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