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lilavasa [31]
3 years ago
10

Which statement about plant classification below is true?

Biology
2 answers:
andrew11 [14]3 years ago
8 0

Answer:

A. No nonvascular plants bear seeds, but not all vascular plants bear seeds

Explanation:

Nonvascular plants are those that do not have conductive vessels (xylem and phloem), in addition, nonvascular plants do not have seeds, fruits or flowers.

Some <u>examples</u> of nonvascular plants are mosses and liverworts.

Most vascular plants (have conductive vessels) have seeds, as in the case of angiosperms and gymnosperms, however, there are also some species of seedless vascular plants, represented by the lycopods and especially the pteridophytes, which are in greater number.

<u>Examples</u> of seedless vascular plants: ferns and horsetails.

SCORPION-xisa [38]3 years ago
6 0

Answer:

B. Not all nonvascular plant bear seeds, but all vascular plants do bear seeds.

Explanation:

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7. Pyrethrums are chemicals used to kill insects. In bed bugs, the mutant-type allele, r, confers resistance, but only in homozy
natita [175]

Answer:

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

Explanation:

It is given that 99% of the bugs were killed after the spray of  pyrethrum. This suggests that 1% of the bugs that were not killed must be homozygous for the mutant type allele "r"

Thus, the frequency of homozygous "rr" species i.e q^2 = 0.01

From this we can evaluate the frequency of mutant "r" allele.

Thus, q = \sqrt{0.01} \\

q = 0.1

As per Hardy-Weinberg first equilibrium equation, p + q = 1

Substituting the value of q in above equation, we get

p = 1 - q\\p = 1 - 0.1\\p = 0.9

Thus, the frequency of homozygous normal bug is equal to

p^2 = 0.9^2 = 0.81

As per Hardy-Weinberg second equilibrium equation-

p^2 + q^2 + 2pq = 1\\

Substituting all the available values we get -

0.81 + 0.01+ 2pq = 1\\2pq = 0.18

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

3 0
3 years ago
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