Now, this time observe what happen if the box where pushed toward each other

likoan [24]

The Answer is A

Explanation:

End hunger, achieve food security and improved nutrition and promote sustainable agriculture.

6 0

2 years ago

HELP ME WITH THIS 2 PLZ

Aliun [14]

The first option for 1 and the second option for 2

7 0

2 years ago

Read 2 more answers

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

E. coli DNA polymerase III is both highly accurate and highly processive. a) Describe how Pol III avoids the incorporation of im

kogti [31]

Answer/Explanation:

DNA polymerase is the enzyme responsible for replicating DNA. It is hugely important that is performs its functions accurately, as if incorrect bases are incorporated this can lead to mutations that disrupt the structure and function of genes. It adds nucleotides in a 5' - 3' direction only.

DNA polymerase III also has high processivity, which means that for every time it binds DNA, it is able to add many bases before it becomes dissociated.

A. DNA polymerase avoids the incorporation of improperly paired nucleotides in two ways:

The first way depends on the structure of the enzyme. If the nucleotide that the enzyme is in the process of adding is not complementary to the template, then the nucleotide will not align with the template, and thus it is more inefficient to add. This inefficiency means the nucleotide is more likely to leave the active site before it is added, and DNA polymerase can replace it with the correct nucleotide.

It also has proofreading capabilities. This means, when an incorrect base is added, it recognises the error and can fix this. It can do this because it possesses 3'-5' exonuclease activity. That means, it can chop out incorrectly added bases.

B. Ribonucleotides are the nucleotides that are incorporated into a growing RNA molecule. They are different from deoxyribonucleotides because of the differences in the sugar backbone (ribose vs deoxyribose). Their incorporation would disrupt the structure and function of the DNA, leading to problems with transcription and replication.

DNA polymerase avoids incorporating these nucleotides primarily because of the structure of the enzyme. Ribonucleotides cannot fit into the active site of DNA polymerase due to what is called a "steric filter" or "steric gate". This gate/filter function is performed by specific amino acid residues which usually have a bulky side chain and thus block the incorporation of the 2'OH of the ribose sugar (which is lacking in the deoxyribose sugar)

9 0

3 years ago

The Hardy-Weinberg law is used to calculate allele and genotype frequencies ____. Question 1 options: in small populations when

Solnce55 [7]

Your answer can be found on quizlet, Brainly won’t let me comment the answer. Just copy+paste :)

7 0

2 years ago