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2 years ago

How would an increase in nest sites affect a population of pigeons?

Biology

1 answer:

Vikki [24]2 years ago

4 0

Answer:

B. The population would increase.

Explanation:

The population would increase because the more spaces for the pigeons to breed and raise their offsprings.

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Who is credited with the discovery of the structure of dna?

frez [133]

James Watson is credited with the discovery is the stricture of DNA.

he gibe the 3D structure of DNA in 1962 ,
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★ hope you like it ★

8 0

3 years ago

Read 2 more answers

If i need 3/4 cup of oats for a batch of muffins how much muffins can i make with 1 1/2 cups of oats

malfutka [58]

Answer:
2 batches of muffins

5 0

2 years ago

13)

Julli [10]

Multiply 5730 years by 2 since two half-lives have gone by for carbon.

Explanation:

The half-life of a radioactive isotope depicts the measure of time that it takes half of the isotope in an example decay. On account of radiocarbon dating, the half-existence of carbon 14 is 5,730 years

The half-life of carbon-14 is 5730 years.

In this manner, after

1 half-life there is 50 % = 1/2 of the first amount left.

2 half-lives there is 25 % = 1/4 of the first amount left.

25% is two half-lives.

Every 50% of life requires 5730 years.

So two half-lives require 2 × 5730

6 0

3 years ago

A(n) ____________ is a family-tree like representation of how organisms are related based on ____________ .

yan [13]

Answer: A PEDIGREE is a family-tree like representation of how organisms are related based on ANCESTRY. It shows genetic relatedness or blood relationship between individuals or organisms.

Explanation: Pedigree can be used to determine

1. Family history or genealogy.

2. Genetic or blood relationship.

3. Coefficient of relationship.

4. Degree of relationship.

5. Proportions of shared genes.

6. Mode of transmission of diseases.

7. Risk evaluation or determination.

Genetic counseling.

8. Linkage analysis.

Pedigree symbols are used in constructing pedigree.

4 0

2 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