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4 years ago

While gutting a fish, Nathan encountered an air-filled sac-like structure. What is the purpose of this structure in fish?

Biology

1 answer:

eimsori [14]4 years ago

4 0

D. It provides buoyancy to a fish.

This air pocket is known as the swim bladder, and also called air bladder. This is a buoyancy organ possessed by most bony fish. The air bladder is located in the body cavity and is derived from an out pocketing of the digestive tube.

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What is the most important factor that determines the rate of diffusion in biological systems?

miv72 [106K]

The factors that determines the rate of diffusion in the biological system include: temperature, size of the particles, diffusion distance and concentration gradients. Of all these factors, the most important one is the TEMPERATURE.
The relationship between temperature and the rate of diffusion is a direct one, that is, the higher the temperature the higher the rate of diffusion and vice versa. At high temperature the molecules in the particles move faster because they have higher amount of energy.

3 0

4 years ago

H2=3.4g+O2=10g —> H2O= ?

tangare [24]

Answer:

mass of 1 mole of H 2 = 3.4 g + O 2 = 10 g → H 2 O = 3614.35999774 g

Explanation:

6 0

3 years ago

Imagine a genetic counselor working with a couple who have just had a child who is suffering from Tay-Sachs disease. Neither par

Alexxx [7]

Answer:

C. "Because you have had one child with Tay-Sachs, you must each carry the allele. Any child you have has a 25% chance of having the disease

Explanation:

Tay-Sachs disease is an autosomal recessive disorder that means a child will acquire the phenotype or we can say disease only when both the alleles which cause disease are present in the child.

Here, in this condition neither parent has Tay-Sachs but they gave birth to a child who is suffering from the disease then it simply means that both the parents are carrier of this disease i.e. they both are heterozygous.

The cross and probability of healthy and diseased child is shown as under:

Mother Father

Parents Tt Tt

/ | | \

Off-springs TT Tt Tt tt

Out of these 4 probable off-springs only 3 with genotype TT, Tt & Tt will be healthy while the 1 with genotype tt will be diseased. So, the probability of diseased child is 1/4 which is 25%.

Also, the birth of children is an independent event so if one child has already been born with this disease then it doesn't mean that the next 3 children will be healthy for sure i.e. birth of first child cannot influence the next child. In short we can also say that in the next birth, there is still 25% chance of the child being born with this disease because previous birth has nothing to do with next birth.

7 0

3 years ago

If data for the growth of this bacterial population continued to be recorded, would the data point at 60 hours be above or below

Anon25 [30]

Answer:

Above 37 million

Explanation:

Explanation in picture below

8 0

3 years ago

An individual heterozygous for eye color, skin color, and number of eyes mates with an individual who is homozygous recessive fo

sukhopar [10]

Answer:

B) 1 black eyes, green skin, two eyes : 1 black eyes, green skin, one eye : 1 black eyes, white skin, two eyes : 1 orange eyes, green skin, two eyes : 1 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye

Explanation:

Parentals: BbGgCc x bbggcc

The punette square diagram of the cross of these two parents is attached.

B = black eyes, b = orange eyes;

G = green skin, g = white skin;

C = two eyes, c = one eye

Progeny with BbGgCc genotype → black eyes, green skin, two eyes = 8

Progeny with BbGgcc genotype → black eyes, green skin, one eye = 8

Progeny with BbggCc genotype → black eyes, white skin, two eyes = 8

Progeny with bbGgCc genotype → orange eyes, green skin, two eyes = 8

Progeny with Bbggcc genotype → black eyes, white skin, one eye = 8

Progeny with bbGgcc genotype → orange eyes, green skin, one eye = 8

Progeny with bbggCc genotype → orange eyes, white skin, two eyes = 8

Progeny with bbggcc genotype → orange eyes, white skin, one eye = 8

So, if we will see them in the form of ratio then each of the genotype will be in the ratio of 1 i.e. black eyes, green skin, two eyes: black eyes, green skin, one eye: black eyes, white skin, two eyes: orange eyes, green skin, two eyes: black eyes, white skin, one eye: orange eyes, green skin, one eye: orange eyes, white skin, two eyes : orange eyes, white skin, one eye = 1:1:1:1:1:1:1:1.

So, in view of above option B is most appropriate.

Note: In a trihybrid cross, a total of 64 progeny types may be produced.

6 0

3 years ago