Answer:
d. 110
Explanation:
<u>Parental cross</u>: Gh/Gh x gH/gH
<u>F1</u>: Gh/gH
<u></u>
<u>The following test cross experiment is done:</u>
Gh/gH x gh/gh
<u>Gametes produced by these individuals:</u>
- gh/gh: gh (probability of 1)
- Gh/gH:
- Gh (parental)
- gH (parental)
- GH (recombinant)
- gh (recombinant)
The formula that relates genetic distance with recombination frequency is:
<h3>
Genetic Distance (m.u.)= Recombination Frequency X 100</h3>
In this problem:
22 m.u. / 100 = Recombination Frequency
0.22 = Recombination Frequency
The recombination frequency altogether is 0.22, but there are 2 possible types of recombinant gametes and when one is generated, the other one is generated as well. Therefore, each recombinant gamete has a frequency of half the total recombination frequency: 0.11
In an offspring of 1000 individuals, I would expect 110 to be GH/gh.
Because whatever grows there, dies there, and the nutrients remain there. Since wetlands are mostly flat with stagnant or slow moving water. The wetness also keeps the nutrients in place, unlike dry places where the nutrients can be blown away or eroded by water
Answer: hail, snow, sleet, rain, graupel, ice pellets and drizzle are the main forms of precipitation.
Answer:
Not appropriate question.
Explanation:
Required Information is not provided to answer the question.
