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Alexandra [31]
3 years ago
11

Why do adult urochordates (tunicates) lack notochords, even though larval urochordates have them? Larvae use notochords to _____

.
A) aid in swimming; adults are sessile and thus no longer propel themselves.
B) stiffen their bodies; in adults, the notochord is replaced by a column of bone.
C) induce tissue differentiation; in adults, tissue is already differentiated.
D) organize their nervous systems; adults' nervous systems are fully developed and do not change.
Biology
1 answer:
kirza4 [7]3 years ago
8 0

Answer: Option A

Explanation:

A tunicate is marine invertebrate animal  which is a part of chordates. They have notochords and dorsal nerve chord. The subphylum of the organism is called as Urochordata.

The adult urochordates lack notochord but the larval stage of the urochordates have notochord. This is because it helps in swimming of the organism which is required only at the larval stage.

The adult stage of the tunicates is sessile and they do not swim so they do not require notochord.

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Prokaryote's ability to regulate patterns of gene expression most likely promotes the organism's survival by ________.a) allowin
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Read 2 more answers
In a species of bird, the allele for the striped trait is recessive to the allele for the speckled trait. Suppose that a coastal
mylen [45]

Answer:

The answer is "0.42"

Explanation:

Please find the complete question in the attached file.

Coastal community striped allele intensity = q = 0.62

Landlocked community, strip intensity allele=q'=0.40

Its eliminated allele frequency for movement = q"=0.42 \ in \ indoorscommunity And now in the sturdy vineyard optimum communities, genotype frequency of stripped characteristic (heterozygous device) in coasts=q_2=0.39

Heritable (striped) allele frecency:

\to q=\sqrt{q_2}

      =\sqrt{0.39} \\\\ =0.624 \\\\ =0.62

(round up to the closest cent) is therefore the only frequency of coastal people.

The genome of pulling function in host population= q'_2=0.16 Heterozygous feature regularly.

Therefore the inland community rate of recessed (striped), allele rate:

\to q'= \sqrt{q'_{2}}= \sqrt{0.16}=0.40

Following migration;

Its percentage of coastal migrants:

m=10\% \\\\

   = \frac{10}{100} \\\\= 0.1

Coastal population Non-immigrant percentage of coastal residents:

=1- m\\\\=1-0.1\\\\ =0.9

Stripping coastal community of allele rate after immigration:

(q")=\text{(stripping of migrant only freq)}\times (m)+\text{(stripping allele baud rate for non-immigrants)} \times ( 1-m)

       = [q \times m]+[[q' \times(1-m)]\\\\=(0.62 \times 0.1)+(0.40 \times 0.9)\\\\=0.062+ 0.36\\\\=0.422\\\\=0.42

8 0
3 years ago
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