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3 years ago

Which end of the water molecule attracts most of the electrons and acts negative?

1 answer:

4 0

A water molecule is a polar molecule because ot the shape of its molecule. One side of the water molecule is positively charged while the other is negatively charged. The oxygen and the hydrogen atom in water shared electrons by covalent bonding, the oxygen atom has a slight negative charge and attract most of the electron to itself.

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There are five different types of nucleotide bases found in living things.

poizon [28]

Answer:

Explanation:

Birds and flowers have different arrangements of the bases.

Nucleotides are termed as organic molecules which serve as monomer units. Their work is to form nucleic acid polymers, for example, ribonucleic acid and deoxyribonucleic acid.

They serve as biomolecules in the life-forms on earth. Nucleotides work as building blocks of nucleic acid. They are composed of the phosphate group, five-carbon sugar, and nitrogenous base.

6 0

3 years ago

Six million years ago, hominids had ____________ compared to present-day hominids.

Inessa [10]

I believe the answer is a

7 0

3 years ago

4) A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality)

docker41 [41]

Answer and Explanation:

A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).

The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?

1st cross:

Parental) grogro ro+ro+ x gro+gro+ roro

F1) gro+gro ro+ro

2nd cross:

Parental) gro+gro ro+ro x grogro roro

Gametes) gro+ro+ gro ro

gro+ro gro ro

gro ro+ gro ro

gro ro gro ro

Punnet square)

gro+ro+ gro+ro gro ro+ gro ro

gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro

F2)

0.518 grogro ro+ro (518 individuals)

0.471 gro+gro roro (471 individuals)

0.006 grogro roro (6 individuals)

0.005 gro+gro ro+ro (5 individuals)

Total number of individuals 1000

Note: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.

To know if two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.

In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.

By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are gro+ro+ and gro ro, while the parental gametes are gro+ro and gro ro+.

So, the genotype, in linked gene format, of the double heterozygote individual in the F1 is gro+ro/gro ro+.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 6 + 5 / 1000

P = 11 / 1000

P = 0.011

The genetic distance between genes, is 0.011 x 100= 1.1 MU.

Genetic Linkage Map:

Parental Phenotypes)

-----gro+------ro---- -----gro------ro+----

----- gro ------ro---- ---- gro------ ro ----

Recombinant phenotypes)

-----gro+------ro+---- -----gro------ro----

----- gro ------ ro---- -----gro------ro----

If the genes were unlinked and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been

4/16 = 1/4 gro+gro ro+ro

4/16 = 1/4 gro+gro roro

4/16 = 1/4 grogro ro+ro

4/16 = 1/4 grogro roro

Their phenotypic frequencies would be 1:1:1:1 related.

7 0

3 years ago

Which of the following best explains what would happen to animal cells if plant cells did not perform photosynthesis?

ololo11 [35]

Answer:

D. Animal cells would not have the materials needed to produce energy in cellular respiration.

Explanation:

6 0

3 years ago

Read 2 more answers

Which products result from carbohydrate metabolism?

a_sh-v [17]

Answer:

pyruvate and acetyl-CoA

Explanation:

The first step of respiration reactions is glycolysis. When glucose is broken down in glycolysis, the first molecule that is produced is pyruvate. If pyruvate continue to aerobic respiration, it must enter the matrix of mitochondria and be oxidised to Acetly Co-A.

6 0

3 years ago