Why do scientists typically include as large a sample size as possible in their experiments?

50 POINTS AND BRAINLEIST What does the Alaskan native name Nauraq mean

ikadub [295]

Answer:

N is for number one, make sure to take care of yourself above all others

A is for attractive, inside and out

U is for upstanding, the honorable way to be

R is for romp, you know how to have fun!

A is for achievements, the many over a lifetime

Q is for qualified, no doubt you are!

Explanation:

How unique is the name Nauraq?

Out of 5,933,561 records in the U.S. Social Security Administration public data, the first name Nauraq was not present. It is possible the name you are searching has less than five occurrences per year.

Weird things about the name Nauraq:

Your name in reverse order is Qaruan. A random rearrangement of the letters in your name (anagram) will give Qrauna hope it helps plz Mark me brainliest

5 0

3 years ago

If research is restricted, then the new discoveries made are limited. Which action is most likely to increase the number of disc

Rasek [7]

The answer is D. Increasing government funding of research

6 0

3 years ago

Which types of organisms have a cell wall and which have a cell membrane.

Ratling [72]

All plants have a cell wall, all mammals have a cell membrane. a cell membrane is semi-permeable it allows mammal cells to absorb needed nutrients, oxygen and deposit waste into the blood stream where it is excreted through urine and feces. a plant cells nutrients are absorbed through the root system and leaves, waste is excreted in the form of oxygen through the plant itself.

6 0

3 years ago

If you cross a mule and a horse, you get a donkey. How many chromosomes does a donkey have?

madreJ [45]

Answer:

62

Explanation:

31 pairs

A mule is the offspring of a male donkey (a jack) and a female horse (a mare). A horse has 64 chromosomes, and a donkey has 62. The mule ends up with 63. Mules can be either male or female, but, because of the odd number of chromosomes, they can't reproduce.

3 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