Growing two species of

As engineers prepare to build a hydroelectric dam, to which factor should they pay special attention

tekilochka [14]

While preparing to build a hydroelectric dam, the safety for various potential hazards must be given attention. Some of these include special attention towards the channels leading to dangerous areas or spillways, attention towards the effectiveness of the barriers, areas where higher currents exist, attention must also be given towards environmental aspects such as damming the rivers impacts wild and aquatic life, reduction of dissolved oxygen harming the fishes, and also while building the dam the communities living around should be taken into consideration that there is no potential hazard to them.

7 0

3 years ago

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The opposite of homeostasis is:

Zanzabum

Answer:

Disequilibrium

Explanation:

Homeostasis is a relatively stable equilibrium

While Disequilibrium is unsteady or unstable

5 0

2 years ago

Read 2 more answers

What type of microscopy would you use to study the detailed internal structure of a mitochondria? A.Light microscopy B.Scanning

dolphi86 [110]

Answer:

C: Transmission Electron Microscopy

Explanation:

Transmission Electron Microscopy shows the complex internal membrane structure of mitochondria

6 0

3 years ago

2. Dominant trait: cleft chin (C) Mother’s gametes: Cc

andre [41]

.2. Offspring Genotypes will be Cc or cc.

Offspring phenotypes : Cleft chin or no cleft chin.

% chance child will have cleft chin: 50%

3. % chance child will have arched feet: 25%

4. % chance child will have blonde hair: 50%

5. % chance child will have normal vision: 25%

Explanation:

CASE 1 :

Dominant trait: cleft chin (C)

Recessive trait: lacks cleft chin (c)

Father’s gametes: cc

Mother’s gametes: Cc

There are two possible combination of Gametes ,

C fom mother and c from father= Cc

c from mother and c from father = cc

Gametes of Cc Parents= $\frac{1}{2}C + \frac{1}{2} c$ ........(i)

Gametes of cc parents = $\frac{1}{2}c + \frac{1}{2}c$ .........(ii)

Combining (i) and (ii) we get,

$\frac{1}{2} Cc + \frac{1}{2} cc$

There fore offspring Genotypes will be Cc or cc

Offspring phenotypes :

Genotype Cc then phenotype= Cleft chin

Genotype cc then phenotype = Lacks cleft chin.

percentage chance child will have cleft chin = $\frac{0.5}{1}$ ×100

Therefore the chance is 50%.

CASE 2 :

Dominant trait: flat feet (A)

Recessive trait: arched feet (a)

Mother’s gametes: Heterozygous (Aa)

Father’s gametes: Heterozygous (Aa)

There are four possible combination of genotypes are =AA , Aa, Aa and aa

i.e. A from mother, A from father= AA

A from mother, a from father =Aa

a from mother, A from Father = Aa

a from mother, a from father = aa

Gametes of Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

Gametes of other Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

..................................................................................

$\frac{1}{4} AA + \frac{1}{4} Aa$

+ $\frac{1}{4} Aa$ + $\frac{1}{4} aa$

..........................................................................................

$\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa$

Offspring Genotypes will be: AA or Aa or aa

Offsprings phenotype will be:

Genotype AA then phenotype will be Flat feet

Genotype Aa then phenotype will be flat feet

Genotype aa then Phenotype will be arched feet.

Percentage chance child will have arched feet = $\frac{0.25}{1}$ × 100 = 25%

CASE 3:

Dominant trait: Brown hair (B)

Recessive trait: Blonde hair (b)

Mother’s gametes: Homozygous recessive (bb)

Father’s gametes: Heterozygous (Bb)

This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.

Resulting in half Bb and halve bb combination.

Genotypes will be Bb or bb

Phenotypes will be :

Genotype Bb then phenotype Brown hair

Phenotype bb then Phenotype bb.

% chance child will have blonde hair: 50%

CASE 4:

Dominant trait: farsightedness (F)

Recessive trait: normal vision (f)

Mother’s gametes: Heterozygous (Ff)

Father’s gametes: Heterozygous (Ff)

This Case is similar to case 2

it will result in one-fourth FF , half Ff and one-fouth ff combination.

Therefore Genotypes will be: FF, Ff and ff

Phenotypes:

Genotype FF then phenotype farsightedness

Genotype Ff then phenotype farsightedness

Genotype ff then phenotype normal vision.

% chance child will have normal vision: 25%

3 0

3 years ago

É correto afirmar que a presença de lagartas em espigas de milho se deve:

slavikrds [6]

Answer:

C - ao desenvolvimento de ovos depositados por borboletas.

Explanation:

Antes de responder a sua pergunta, eu preciso te explicar que esse não é o servidor brasileiro e sim o servidor dos EUA, você deve ter percebido que toda a linguagem do site esta em inglês. Isso significa que você só deve postar perguntas em português no campo "World Languagens" mesmo que a sua pergunta seja de biologia como essa. Caso você não faça isso, a sua pergunta pdoe ser excluída, sem ser respondida.

As lagartas que surgem nas espigas de milho, são resultado do ciclo de vida da lagarta e refere-se ao momento em que os ovos depositados nas plantas de milho pelas borboletas se chocam, liberando as lagartas, que se desenvolverão e virarão novas borboletas.

É impossivel que um grão e milho vire uma lagarta, ou que o sabubo apodrecido pode gerar lagartas. Além disso, o ´processo de geração espontanea é uma teoria cientifica invalida e que não deve ser considerada.

8 0

3 years ago