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lara31 [8.8K]
3 years ago
8

PLEASE HELP! 30 POINTS!

Biology
2 answers:
Dmitriy789 [7]3 years ago
5 0
<h2>Answer:</h2>

The correct options are 1,4 and 5. By looking at phylogenetic tree we can conclude this;

1. Species A and C may have shared features from a common ancestor.

4. Species B and E may have shared features from a common ancestor.

5. Species A is more related to species B than species B is related to species C

<h2 />
Sonja [21]3 years ago
5 0

Answer:

Option 1, 4 and 5

Explanation:

In a phylogenetic tree, the species which emerge from the same node are more closely related as compared to the other parallel species originating from the same ancestors but at a different node.

Hence, species A is more closely related to species B as compared to the relation between B and C. Option 5 is correct

Both the species B and E will share some characteristics as they have evolved from the same ancestors, though they do not emerge from the same node. Thus, option 4 is also correct.

Similarly A and C have some common characteristics. Thus, option 1 is correct

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The cell membrane of a red blood cell will allow water, oxygen, carbon dioxide, and glucose to pass through. Since other substan
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Answer:

The correct answer would be B. semi-permeable.

Plasma membrane of any cell including red blood cell is semi-permeable in nature as it is made up of lipid bilayer.

It permits the passage of certain substances across the membrane while it blocks other substances.

Gases, small uncharged or non-polar substances such as oxygen, carbon dioxide, et cetera can easily pass through the cell membrane of a cell.

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2 years ago
A locus that affects susceptibility to a degenerative brain disease has two alleles, V and v. In a population, 16 people have ge
Pie

Answer:

Yes, the population is evolving

Explanation:

Hardy-Weinberg equilibrium states that a population is not evolving if it is in hardy weinberg equilibrium as the alleles frequency do not change over time in generations.

In mathematical equation, the genotype frequency will be -P²+2Pq+q²=1 and allele frequency will be p+q=1, where p = dominant allele and q will be recessive allele.

In the question, since the observed genotype are-

P² - VV= 16

2Pq- Vv- 92

q²- vv- 12

<u>1. Calculate allele frequency</u>

Now we will calculate if the expected value is same or close or not.

We will calculate allele frequency

Population has 120 individuals which means it has - 240 alleles.

In which V (p) will be- 32 +92= 124 allele

p- dominant allele/ total allele=  124/ 240 = 0.52

P= 0.52

q will be 1-0.52= 0.48

<u>2. Expected genotype frequency</u>

Now we will calculate expected genotypic frequency.

P²- 0.52 X 0.52= 0.27

2Pq = 2 x 0.52x 0.48 =  0.5

q²= 0.48 x 0.48 =  0.23

<u>3. Expected progeny </u>

P²= 0.27 x 120= 32 VV

2Pq= 0.5 X 120= 60 Vv

q²= 0.23 x 120= 28 vv

Since the observed value is not equal to the expected value therefore the population is not in Hardy- Weinberg equilibrium and the population is evolving.

Thus, Yes, the population is evolving is correct.

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