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Novay_Z [31]
3 years ago
13

A potato plant can reproduce asexually so that one potato can often give rise to many other potatoes. Strawberry plants can repr

oduce sexually.
A potato plant puts a large amount of its energy into growing a tuber, which is the part that humans eat. The tuber can grow large and provide energy for the plant during the winter, or it can be used to grow new potato plants. A strawberry plant, on the other hand, puts much of its energy into growing flowers to attract insects that aid in pollination. After fertilization, the strawberry plant must then also grow seeds as well as fruit to attract animals that disperse the seeds.

Based on this information, the asexually reproducing potato plant most likely devotes _______ to reproduction than the sexually reproducing strawberry plant.
A.
fewer spores
B.
more energy
C.
less energy
D.
more seeds
Biology
2 answers:
valentinak56 [21]3 years ago
7 0
C. Less energy (hope this helps)
notsponge [240]3 years ago
6 0

Answer:

C. Less Energy

Explanation:

The potato plant only has to grow its main part a tuber which can help grow more multiple parts which will easliy create more offspring. a strawberrie  on the other hand needs to grow its main fruit grow flowers and seeds in itself to continue its reproduction cycle. The strawberry also takes a longer time to be able to reproduce.

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In a population of plants, the allele for long stems (S) is completely dominant over the allele for short stems (s). If 35% of t
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Answer:

The correct answer would be 48 percent.

It can be solved by using the Hardy-Weinberg equation.

p² +  2pq+ q²  = 1

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where, p² represents homozygous dominant's frequency that is, SS in this case,

q² represents homozygous recessive's frequency that is, ss in this case,

2pq represents heterozygous dominant's frequency that is, Ss in this case.

p represents dominant allele's frequency and q represents the recessive allele's frequency.

Now, the percentage of homozygous recessive or short stems is 35%. Therefore, the frequency of homozygous recessive would be 0.35.

Thus, the value of q² = 0.35

So, q = \sqrt{0.35} = 0.59 which is approximately equals to 0.6.

As p + q = 1 so, q = 1 - 0.6 = 0.4

The frequency of heterozygous (Ss) = 2pq = 2(0.6)(0.4) = 0.48

Hence, the frequency of heterozygous would be 48 percent.

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