All categories

3 years ago

If an organism has 6 haploid chromosomes, how many chromosomes are present? 6 12

1 answer:

4 0

The Haploid number is half (haploid cells are sex cells egg and sperm they have half as many chromosomes as the diploid cells which are normal cells ) of the chromosomes so that would be 12. To clarify that if a organism has 12 chromosomes and reproduce sexually it takes two
parents each contributing ether a egg or sperm each egg or sperm contains half of the genetic
 material

Your answer would be (6)

Hope this helps.

You might be interested in

Which organisms transform nitrogen to a form that is useful to plants

Natalka [10]

Bacteria

In the process of the nitrogen cycle. 
The nitrogen cycle is a biogeochemical succession process of nitrogen that involves: fixation, ammonification, nitrification, and denitrification. Like any other biogeochemical cycles. This process undergoes and affects the biological, geometrical and chemical aspects in the ecosystem and the abiotic and biotic community. Hence, the nitrogen cycle leads the abiotic component –nitrogen- to contribute to the biotic community, decomposition and primal production. Further, it becomes an essential part of the environment because some life components are contains it, similarly, amino acids, nucleic acids in RNA and DNA.

3 0

3 years ago

Read 2 more answers

Green olives may be preserved in brine, which is a 20-30% salt solution. How does this method prevent contamination by microorga

vredina [299]

The correct answer is Bacteria can't survive in a hypertonic solution because they lose water.

The 20-30% salt solution is hypertonic as Compared to the bacterial concentration of salt. In this case, the bacteria present in the solution would lose water. Due to excess loss of water, the bacteria would eventually shrink and die. Hence, green olives preserved in the brine would be safe from microbial attack.

7 0

3 years ago

HELP Name three carbohydrates that contain glucose as a monomer.

Tju [1.3M]

Answer:

Carbohydrates are classified into three subtypes: monosaccharides, disaccharides, and polysaccharides.

Explanation:

7 0

3 years ago

Make connections Which chemical group is most likely to be responsible for an organic molecule behaving as a base (see Concept 3

spin [16.1K]

Answer:

The correct answer will be option-C.

Explanation:

The functional groups are the group of atoms which gets attached to the skeleton of carbon in a molecule and provides certain characteristics to the molecule. The main functional group includes hydroxyl, carboxyl, amino, phosphate, methyl carbonyl and sulfhydryl.

These groups could be acidic or basic depending upon them if their structure will be able to take H⁺ ions from solution or donate H⁺ ions. In the given question, the basic group among these is an amino group which has the ability to accept H⁺ ions from the solution and form NH₃⁺ ions.

Thus, Option-C is the correct answer.

4 0

3 years ago

Hexokinase can use glucose or fructose as a substrate. The Km forglucose is 0.15 mM and for fructose Km = 1.5 mM.

babymother [125]

Answer:

a) glucose because Km is less for glucose.

b) For glucose, V as a percentage of Vmax = 50 %

For fructose, V as a percentage of Vmax = 9 %

Explanation:

a) Hexokinase has highest affinity for glucose because it's Km for glucose is low.

When enzyme [E] binds a substrate [S] an [ES] complex is formed. It is represent as under:

[E] + [S] ⇌ [ES] ⇌ [E] + [P]

As soon as [ES] is formed it may either form [E] & [P] in a forward reaction or dissociate back into [E] & [S]. Km is known as dissociation constant which represents the extent to which [E] & [S] are being formed. A higher value of Km represents formation of more [E] & [S] back from [ES]. It also means that binding affinity of enzyme with substrate is less or we can say that Km is inversely proportional to binding affinity.

Km ∝ 1 / binding affinity.

So, it simply means that lower the Km more will be the affinity of the enzyme for the substrate. In the given question, Km for glucose is 0.15 mM which is less than fructose (Km for fructose = 1.5 mM) that is why glucose is the answer.

b) The formula for Michaelis–Menten equation is as under:

V = Vmax × [S] / ([S] + Km)

V / Vmax = [S] / ([S] + Km)

Here, for glucose Km = 0.15 mM and [S] = 0.15mM.

So V as a percentage of Vmax for glucose will be as under:

V / Vmax = 0.15/0.15 + 0.15

→ V / Vmax = 0.15/0.30

→ V / Vmax = 0.5 = 1/2 = 50 %

Here, for fructose Km = 1.5 mM and [S] = 0.15mM.

So V as a percentage of Vmax for fructose will be as under:

V / Vmax = 0.15/0.15 + 1.5

→ V / Vmax = 0.15/1.65

→ V / Vmax = 0.09 = 9/100 = 9 %

5 0

3 years ago