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sergij07 [2.7K]
3 years ago
5

If correct i will mark the brainliest if can

Biology
1 answer:
Dmitrij [34]3 years ago
8 0

Answer:

the answer is Gelisols

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Investigators should keep in mind that eyewitness identification is highly infallible. true false
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Investigators should keep in mind that eyewitness identification is highly infallible. true, since the law takes the identification as a positive test and completely commits those involved and an error in it could tarnish the course of justice and consequently of the investigators.
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Is rice pudding a homogeneous or heterogeneous mixture?
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<span>Rice pudding would be an example of a heterogeneous mixture. A heterogeneous mixture means that different constituents of a mixture can be identified. Rice pudding consists of rice suspended in a cream. It is possible to identify the grains of rice and the creamy part of the mixture. A homogeneous mixture looks the same throughout--you cannot identify individual parts or ingredients. The parts of a homogeneous mixture are said to be uniformly distributed. Rainwater and air are examples of homogeneous mixtures.</span>
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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 =<u> </u>\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

                                       <u>..................................................................................</u>

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

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

                                   <u>..........................................................................................</u>

                                <u>\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa</u>

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%

 

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Imagine an island covered in forests that has a population with deer that have no predators.
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1. Limited energy supply to other organisms
2. Herbivores will be scarce because they don’t mate
3. Avoid pollution
Adding sanctions and punishing offenders
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