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sladkih [1.3K]
3 years ago
7

What were Punnett squares developed to do?

Biology
1 answer:
WITCHER [35]3 years ago
7 0

Answer: to predict the probability of offspring genotypes

Explanation: Punnett squares are developed to predict the probability of offspring genotype. Punnett square is a square shaped diagram used for predicting the possible genotypes of a cross between two organisms. Punnett square is named after Reginald C. Punnett, who invented the approach. In punnett squares, the genotypes of the parents are written on the top and left hand side of the square while the possible genotypes of their offsprings are written in the boxes.

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In which animals are the blood and the interstitial fluid considered to be the same body fluid?
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Answer:

In insects such as cockroach, grasshopper

Explanation:

Insects such as grasshopper has open circulatory, in them the body fluid is the hemolymph. The hemolymph does contain any respiratory pigment. It bathes the organs directly as it is pumped between open spaces in the body.

Thus, in insects, the blood and the interstitial fluid considered to be the same body fluid?

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if there is no oxygen during the cellular respiration process, the Krebs cycle does not occur. True or False
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Temperatures in ponds and lakes vary by season.
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1)What two processes fuel the carbon cycle?
never [62]

Answer:

In the natural carbon cycle, there are two main processes which occur: photosynthesis and metabolism. During photosynthesis, plants use carbon dioxide and produce oxygen.

Explanation:

4 0
2 years ago
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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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