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Licemer1 [7]
2 years ago
10

How do nocturnal animals see better in the dark than we can?​

Biology
1 answer:
zvonat [6]2 years ago
4 0

Nocturnal animals possess a pathway through which rods connect to the same “dark” circuits used by cones, which allows them better perception of edges, movement, and silhouettes in dim light. Even the nuclei of the rod cells are adapted for night vision. ... One can only guess at what nocturnal animals see.

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If the atomic number of an element is 6 and its mass is 14 how many neutrons are contained in the nucleus?
Goryan [66]
Your answer would be D. You can find the neutrons by subtracting the mass by the atomic number. 14-6=8 neutrons.
3 0
3 years ago
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One way to protect biodiversity is to set aside more wilderness land. Humans are not allowed in these areas.
svlad2 [7]

The answer is going to be false

6 0
3 years ago
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Which cross shows both parents having two different alleles for each of two different genes?
shtirl [24]

Answer:

The correct option is Number 1. AaBb x AaBb

Explanation:

Genes can be described as basic hereditary units. The genes are made up of DNA.

A gene carries two alleles for a particular trait. These alleles are present at the same loci in the homologous chromosomes. If the two alleles are similar then they are termed as homozygous e.g AA, BB. If the alleles for a trait are different then they are termed as heterozygous e.g Aa and Bb.

7 0
3 years ago
Explain why this product of genetic engineering is called recombinant DNA
Alborosie

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6 0
2 years ago
We are studying 3 strains of bacteria, with populations p1, p2, p3, in an environment with three food sources, A, B, C. In a day
Nastasia [14]

Answer:

The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

Explanation:

From the given information:

For  food source A; we have:

3P₁ + P₂ + 2P₃ = 58    units of food A ---- (1)

For food source B; we have:

2P₁ + 4P₂ + 2P₃ = 70   units of food B  ---- (2)

For food source C; we have:

P₁ + P₂  = 20   units of food C    ----- (3)

From equation (1) and (2); we have:

3P₁ + P₂ + 2P₃ = 58

2P₁ + 4P₂ + 2P₃ = 70

By elimination method

 3P₁ + P₂ + 2P₃ = 58

-

 2P₁ + 4P₂ + 2P₃ = 70

<u>                                       </u>

<u> P₁  -   3P₂   + 0    = - 12    </u>

P₁ = -12 + 3P₂   ---- (4)

Replace, the value of P₁  in (4) into equation (3)

P₁ + P₂  = 20

-12 + 3P₂ + P₂  = 20

4P₂ = 20 + 12

4P₂ = 32

P₂ = 32/4

P₂ = 8

From equation (3) again;

P₁ + P₂  = 20

P₁ + 8 = 20

P₁  = 20 - 8

P₁  = 12

To find P₃;  replace the value of P₁ and P₂ into (1)

3P₁ + P₂ + 2P₃ = 58

3(12) + 8 + 2P₃ = 58

36 + 8 + 2P₃ = 58

2P₃ = 58 - 36 -8

2P₃ = 14

P₃ = 14/2

P₃ =  7

Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

4 0
3 years ago
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