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il63 [147K]
3 years ago
13

14. All the animals in Jurassic Park were_____ to control the populations.

Biology
1 answer:
Aleonysh [2.5K]3 years ago
7 0

Answer: Female

Explanation:

Females cant mate but somehow “life found a way”

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The three horizons in Figure 7-1 make up a soil ____________________.
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Multiple soil horizons make up a soil profile. Soil horizons are distinguishable regions of soil. They run parallel to the surface and have different characteristics. The soil horizons include the A horizon, E horizon, B horizon and C horizons.
The soil profile is a vertical cross section showing these horizons.
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3 years ago
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Which o the following is true about cellular resporation
disa [49]
Cellular respiration is the process of synthesizing cellular energy (ATP) from organic sources such as water, glucose and oxygen. These substances are processed through a series of steps in order to produce ATP. The main organelle responsible for ATP synthesis is the mitochondria. 

<span>In plants, the photosynthetic pigment, chlorophyll, is found within chloroplasts. The process of photosynthesis is driven by light and carbon dioxide from the environment, converting these into glucose, water and energy. </span>

<span>photosynthesis occurs in chloroplasts </span>
<span>cellular respiration occurs in mitochondria</span>
5 0
3 years ago
What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent as
ASHA 777 [7]

Answer:

(a) AABBCC × aabbcc->AaBbCc     → 1

(b) AABbCc × AaBbCc->AAbbCC   → 1/32

(c) AaBbCc × AaBbCc->AaBbCc     → 1/8

(d) aaBbCC × AABbcc->AaBbCc    → 1/2

Explanation:

In such cases we calculate the probability of each allelic pair separately.

(a) AABBCC × aabbcc -> AaBbCc  

Aa  Aa   Aa  Aa           Bb  Bb  Bb  Bb         Cc  Cc  Cc  Cc

↓                                   ↓                                ↓  

4/4 = 1                          4/4 = 1                       4/4 = 1    

In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.

In case of gene B, out of the 4 probable allelic combinations, 4 will be Bb so the probability is 4/4 = 1.

In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.

So, the total probability of getting AaBbCc   = 1 x 1 x 1 = 1.                

(b) AABbCc × AaBbCc -> AA bb CC

AA  AA   Aa  Aa           BB  Bb  Bb  bb         CC  Cc  Cc  cc

↓                                                        ↓           ↓  

2/4                                                    1/4          1/4

In case of gene A, out of the 4 probable allelic combinations, 2 will be AA so the probability is 2/4.

In case of gene B, out of the 4 probable allelic combinations, 1 will be bb so the probability is 1/4.

In case of gene C, out of the 4 probable allelic combinations, 1 will be CC so the probability is 1/4.

So, the total probability of getting AA bb CC = 2/4 x 1/4 x 1/4 = 1/32.

(c) AaBbCc × AaBbCc -> AaBbCc

AA  Aa   Aa  aa           BB  Bb  Bb  bb         CC  Cc  Cc  cc

        ↓                                  ↓                               ↓  

       2/4                               2/4                           2/4

In case of gene A, out of the 4 probable allelic combinations, 2 will be Aa so the probability is 2/4.

In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.

In case of gene C, out of the 4 probable allelic combinations, 2 will be Cc so the probability is 2/4.

So, the total probability of getting AaBbCc = 2/4 x 2/4 x 2/4 = 1/8.

(d) aaBbCC × AABbcc->AaBbCc

Aa  Aa   Aa  Aa               BB  Bb  Bb  bb          Cc  Cc  Cc  Cc

      ↓                                        ↓                                ↓  

4/4 = 1                                       2/4                          4/4 = 1    

In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.

In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.

In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.

So, the total probability of getting AaBbCc = 1 x 2/4 x 1 = 1/2.

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3 years ago
Ways to prevent stomach diseases?
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Answer:

<em>Ways to prevent stomach diseases</em><em> </em><em>are</em><em> </em><em>as</em><em> </em><em>given</em><em> </em><em>below</em><em>:</em><em>-</em>

<em><u>Chew food thoroughly, and don't overeat.</u></em>

<em><u>Avoid raw shellfish if you're not sure the source is a safe one.</u></em>

<em><u>Limit your intake of fats and alcohol.</u></em>

<em><u>Get plenty of fluids</u></em><em><u>Exercise daily.</u></em>

<em><u>Exercise daily.</u></em>

<em><u>Avoid foods that cause gas.</u></em>

<em><u>Avoid sweeteners that cause gas such as fructose and </u></em><em><u>surbiton</u></em><em><u> </u></em>

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3 years ago
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