All categories

2 years ago

Sponge larvae are more animal-like than adult sponges because

2 answers:

5 0

Sponges are organisms that are members of phylum prolifera, they are simple multicellular animals without a mouth, muscles, heart or brain.Sponges's tissues contains collar cells and amoebaocytes in its tissues. They feed by drawing water in through their pores and filter out food particles. Sponge larvae are more animal-like than adult sponges because they are free-swimming organisms.

IRISSAK [1]2 years ago

5 0

<span>Sponge larvae are more animal-like because these larvae are free-swimming organisms. These sponges have no nervous systems, digestive systems or any circulatory systems but they depend on the constant water flow through the bodies. Their reproduction is done through both sexually and asexually.</span>

You might be interested in

⚠️❗PLEASE HELP ❗⚠️

Y_Kistochka [10]

the correct answer is Evaporation

hope this helps :)

7 0

2 years ago

Read 2 more answers

Why is it not acceptable to edit the genome of human embryos to treat genetic diseases ? Please answer this !! Will mark Brainli

aleksandrvk [35]

its illegal in a lot of countries

4 0

3 years ago

Read 2 more answers

In a hardy-weinberg population with two alleles, a and a, that are in equilibrium, the frequency of the allele a is 0. 3. What i

zavuch27 [327]

Here, given-

homozygous alleles 'a' have a frequency of 0.3.

Also the alleles are in equilibrium in a Hardy-Weinberg population. The frequency of individuals that are homozygous for this allele are= 0.49.

The Hardy-Weinberg equilibrium can be defined as the principle which states that the variation in the genetic makeup of a population remains constant and unchanged till there are no external interferences, influencing the population.

Calculation-

$p + q = 1\\0.3 + q= 1\\q= 1- 0.3\\\\q= 0.7$

Then to find the frequency of the individuals homozygous for this allele the following formula needs to be used-

$p^{2} + 2pq + q^{2} =1$

$p^{2} = dominant homozygous frequency\\2pq= heterozygous frequency\\q^{2} = recessive homozygous frequency\\$

Thus, the individuals homozygous for the allele can be calculated by $q^{2} = (0.7)^{2} \\= 0.49$

Learn more about the Hardy-Weinberg equilibrium here-

brainly.com/question/16823644

#SPJ4

4 0

2 years ago

Although the foxp2 amino acid sequence is highly conserved, humans have two unique amino acids that chimpanzees don't have. thes

PolarNik [594]

The answer is language. Humans have a high degree of language acquisition and complexity compared to chimpanzees due to foxp2 proteins. Foxp2 are transcription factors that are significant in cell proliferation. The difference of 2 in 700 amino acids in the sequence makes the protein to be differentially regulated between the two species.

4 0

3 years ago

Read 2 more answers

In monkeys, being playful (P) is dominant to being paranoid (p). A heterozygous female is crossed with a homozygous dominant mal

Sergio039 [100]

Answer:

Pp or PP

Explanation:

Being heterozygous mean you have one recessive gene(a), one dominate (A). Homozygous dominant means you have only dominate(AA). if you use a Punnett square, it would look like this:

P P

P PP PP

p Pp Pp

This shows only Pp, and PP are the possible genotypes.

7 0

2 years ago