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alex41 [277]
4 years ago
7

6. A scientist measures the circumference of acorns in a population of oak trees

Biology
1 answer:
melisa1 [442]4 years ago
3 0

Answer:

6. 2 cm

8. Greater than 2 cm and less than 2 cm

Explanation:

Complete Question

6. A scientist measures the circumference of acorns in a population of oak trees and discovers that the most common circumference is 2 cm. What would you expect the most common circumference(s) to be after 10 generations of stabilizing selection?

-2 cm

-Greater than 2 cm or less than 2 cm

-Greater than 2 cm and less than 2 cm

-Can't tell from the information given

8. Refer to question 6, but this time answer what you would expect after 10

generations of directional selection.

-2 cm

-Greater than 2 cm or less than 2 cm

-Greater than 2 cm and less than 2 cm

-Can't tell from the information given

Solution -

6. In an stabilizing selection, the average trait of a given population is favored over extreme values. Hence, it can be said that one specific trait is over presented in a given population. For example – uniform human birth weight, number of eggs a bird lay etc.  

Here in this case if most common circumference is 2 cm , then the tree population will try to have this trait as a stabilizing trait.  

Thus, option A is correct

8. In a  directional selection, the genetic variation decreases as per the environmental change. And hence the average trait will not be selected. Instead traits either higher or lower the average values will be preferred.  

Thus, option C is correct

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The domestic cat genome contains 2.9×109 base pairs. The length of linker DNA in mammals is 50 base pairs. Approximately how man
IgorLugansk [536]

Answer:

Number of nucleosomes in 2.9 * 10^9bp is equal to 1.47 * 10^7

Explanation:

For wounding one nucleosome, total length of DNA required is equal to 146 bp

The length of  linker DNA in mammals is equal to 50 bp

Thus , the total length of DNA that confides between two nucleosome is equal to the sum of wounding length of DNA and the linker length

= 146 + 50\\= 196bp

Thus, in 196bp length of DNA, the total number of nucleosomes is equal to 1

Thus, number of nucleosomes in 2.9 * 10^9bp is equal to

\frac{2.9* 10^9}{196} \\1.47 * 10^7

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