In a population of gerbils, long hair (h) is completely dominant over short hair (h). if 18% of the population has short hair, calculate the percentage of the population that is expected to be heterozygous (hh).
a) 29%
b) 49%
c) 58%
d) 80%

b) 49%

Explanation:

To answer this question, we must assume that the population of gerbils is in Hardy-Weinberg equilibrium.

According to this law, the frequencies of the genotypes will be:

Where p is the frequency of the H allele and q is the frequency of the h allele.

p+q=1

If 18% of the population has short hair, the frequency of the hh genotype is 0.18.

Therefore:

The expected Hh population is 2pq:

The percentage of the population expected to be heterozygous is 0.49x100=49%

the real answer is b. scintillation counter

events in order:

1.       vishnu schist

2. zoroaster granite

3. bass limestone

4. hakatar   shale

5. igneous intrusion

6. shinumo quartzite

7. dox sandstone

8. tapeats limestone

9. bright angel shale

10. muav limestone

11. temple butte limestone

12. redwall limestone  

nonconformity

answer: the unkar group shows angular unconformity because the layers are turned at angle.

unconformity

answer: specifically the angular unconformity occurs between the cambrian and precambrian rocks. the dox sandstone, shinumo quartzite, hakatar shale, and bass limestone have been filled. an angular unconformity occurs between the cambrian and precambrian rocks. the dox sandstone, shinumo quartzite, hakatar shale, and bass limestone have been filled.

disconformity

answer: the disconformities aremarked irregular lines between the sedimentaryrocks layers.