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3 years ago

Nick hypothesizes that wax has a higher melting point than chocolate. How can Nick test his hypothesis?

Biology

2 answers:

Reil [10]3 years ago

5 0

Melting them both in the same environment and measuring the temperature at which each melts

NISA [10]3 years ago

4 0

Answer:

Hypothesis may be defined as the proposed explanation of the particular phenomena. Any statement in the science can be made hypothesis only if it is tested by the scientific experiments.

The nick hypothesis can be tested by measuring the melting point of the wax and chocolate. The similar conditions are used for measuring melting point of both wax and chocolate. The substance that melts early than the other substance has lower melting point. The chocolate melts early than wax and thus wax has higher melting point than wax.

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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 = $\frac{1}{2}c + \frac{1}{2}c$ .........(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

When disinfecting a whirlpool foot spa after use by a client, you must circulate the disinfectant for ______________ or the leng

SIZIF [17.4K]

Answer:

10 minutes

Explanation:

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3 years ago

By controlling the spread of vegetation and the balance of certain fish species, the American alligator serves as a(n) in its ec

seraphim [82]

Answer: Below

Explanation:

Alligators play an important role in maintaining ecosystem balance. Sitting at the top of the food chain, alligators are top predators and help keep other animal populations in balance. By digging holes and leaving trails throughout marshes, they create habitats for fish and marine invertebrates

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2 years ago

What is post-combustion carbon capture?

Olenka [21]

Answer:

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Explanation:

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3 years ago

The _____ consists of protoplasm and a nucleus. A. vacuole B. amoeba C. bacterium D. pseudopodium

malfutka [58]

The correct answer would be amoeba. So the amoeba consists of protoplasm and a nucleus. For an amoeba the nucleus is a major organelle, which is located centrally. As such a major organelle it naturally has a huge role, and that is that it controls reproduction (it contains the chromosomes) and many other important functions (including eating and growth).

4 0

3 years ago