5. There are several schools of thought regarding the most appropriate method fo

Question

5. There are several schools of thought regarding the most appropriate method for reconstruction of phylogenies. Identify the approach that emphasizes overall similarities and the reason it is preferred by some systematists.

a. Cladistics, because it reduces subjectivity.

b. Cladistics, because it focuses on shared derived characters.

c. Phenetics, because it rarely mistakes convergent evolution for recent common ancestry.

d. Phenetics, because it reduces subjectivity.

*6. Internal fertilization influences the evolution of parental care strategies because:

a. Externally fertilizing species do not show parental care.

b. The placement of sperm inside the female predisposes the male to care for the resulting offspring.

c. If postzygotic care is necessary for success, the female is in the position of “last custody”.

d. “Advanced” organisms always have internal fertilization.

11. Honeybees and ants posed a special dilemma to Darwin because:

a. They are so small that he could not experiment on them.

b. They show apparently self-sacrificial traits, which conflicts with evolution by natural selection.

c. They show apparently self-sacrificial traits, which conflicts with haplodiploidy.

d. The females are larger and more powerful than the males, which conflicts with sexual selection.

e. They could survive insecticide, and so were not subject to natural selection.

13. If the half-life of carbon-14 is about 5,730 years, then a fossil that has 2.5 grams carbon-

      14 and 7.5 grams of its stable decay product should be about how many years old?

            a. 22,900

            b. 2,800

            c. 17,200

            d. 11,500

            e. 1,400

Explanation / Answer

5. Answer b. Cladistics uses shared derived characters to point to a common ancestor.

6. Answer c.

11. b. Even though the worker bees are sterile, it is now shown that they also contribute to the evolution of the species.

13. d.

The total weight of the fossil is 10 g. of this 2.5 g is C- 14 and 7.5 g is stable decay product.

Therefore we can conclude that 25% of the original C-14 is still present in the sample.

At the moment of death, the total amount of C-14 in the organism = 100%

Therefore 100 %-------à 50 %-------à 25% - therefore 5730 x 2 for the C- 14 has completely decayed 2 times .

5730 x 2 = 11460. So as option d is the closest answer, we chose option D.

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