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mrs_skeptik [129]
2 years ago
7

A pioneer species is...

Biology
1 answer:
ivanzaharov [21]2 years ago
4 0

Answer:

b

Explanation:

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c. Suppose a large population of turtles consists only of two species, snapping turtle and box turtle, but the percentage of the
vodka [1.7K]

Answer:

D_m=1/2      

Explanation:

From the question we are told that

2 species

Assuming that questions a and be are solved

Diversity index D=2p_1p_2

with p_1 +p_2 =1

Mathematically

D=2p_1(1-p_1)

D=2(p_1-p_1^2)

D=2({\frac{1}{4} -(p_1-1/2)^2}

Therefore

D=\frac{1}{2} -2(p_1-1/2)^2

Giving

D_m=1/2      

p_1=1/2

6 0
3 years ago
Will give brainliest
ryzh [129]

Answer:

the bottom left one is the one

6 0
3 years ago
Read 2 more answers
Help! <br> How many dolphins are usually in a pod??<br> 25 Pointss!
Andrew [12]

Answer: Most pods contain anywhere from 2 – 30 dolphins depending on the species and the situation, however there are occasions when pods gather with other pods to form superpods of 100 or even several thousand dolphins!

Explanation:

7 0
2 years ago
Read 2 more answers
The ecosystem is the lowest level of the biosphere hierarchy to include _______..
netineya [11]
<span>The ecosystem is the lowest level of the biosphere hierarchy to include weather patterns and soil conditions. in the whole biosphere hierarchy, ecosystem actually placed as the second highest (1 place below biosphere) But there are only 2 hierarchy that include weather and soils, and those hierarchies are biosphere and ecosystem. So technically the ecosystem will be the lowest in this matter</span>
8 0
3 years ago
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Alkaptoneuria is a recessive genetic disorder found among North American populations at a frequency of about 1 in 500,000 births
Dafna11 [192]

Answer:

The correct answer is - 1/41,493

Explanation:

Let assume the frequency of the two possible same allele genotype (dominant and recessive) in an inbred population is p and q. Then the frequency of heterozygotes (H) is denoted as:

2pq + 2pqF.  ( where F is the inbreeding coefficient).

The frequency of the two different hoozygotes in inbred population can be calculated as:

p2 + pqF and q2 + pqF. (Where p and q are the allele frequency of the dominant and recessive phenotype.

Given: Frequency of Alkaptonuria (q 2) = 1:500, 000

=> q = 1/707

p = 706/707 ( Approx values)

solution:

Inbreeding coefficient (F) = 1/64

Therefore,

Frequency of Alkaptonuria in second cousins= q 2 + pqF

= 1/500, 000 + (706/707 x 1/707) x (1/64)

= 1/500, 000 + 1/45, 248

= 1/41,493 (approx)

5 0
3 years ago
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