All categories

2 years ago

Which is true regarding the process of meiosis I?

Biology

1 answer:

sashaice [31]2 years ago

7 0

Answer: The correct option is A (chromosome Number is reduced from diploid to haploid)

Explanation:

In sexual reproduction, offspring are produced by fusion of two different sex cells which usually come from two different parents. These sex cells are know as gametes.

Meiosis is a type of cell division that gives rise to gametes in which the chromosome number is halved. Thus, the gamete cell is said to contain a haploid number (n) of chromosomes.

A diploid cell has two sets of chromosomes: one from the male parent and the other from the female parent. When a diploid cell undergoes meiosis the following occurs:

--> the chromosomes replicate once and

--> the nucleus and cell duplicate ( divide equally) twice.

As a result, the diploid parent cell gives rise to four haploid gamete cells. Therefore the statement "chromosome Number is reduced from diploid to haploid ' is true concerning meiosis I.

You might be interested in

the increase to heart rate is a normal homeostatic response and allows for an adjustment of the body's set point during exercise

hodyreva [135]

Answer:

it increases oxygen levels and decrease carbon dioxide levels

3 0

3 years ago

Compare and constraint negative and positive feedback mechanism and explain how they relate to homeostasis

gavmur [86]

Explanation:

Positive feedback occurs to increase the change or output: the result of a reaction is amplified to make it occur more quickly. Negative feedback occurs to reduce the change or output: the result of a reaction is reduced to bring the system back to a stable state.

6 0

2 years ago

How does prokaryotic transcription and translation differ from these processes in eukaryotic cells?

daser333 [38]

Explanation:

The genome of prokaryotes has no introns hence their mRNA does not need splicing like in eukaryotic cells. Also, because the genome of prokaryotes is not delimited from the cytoplasm by a nuclear membrane, ribosomes can attach to the elongating mRNA during transcription and begin translation. Therefore translation of mRNA occurs concurrently with transcription which cannot happen with eukaryotic cells.

In the nucleus of eukaryotic cells, transcription results to a nascent mRNA which is spliced into a mature mRNA.The mature mRNA has to travel outside the nucleus to the cytoplasm to be translated by ribosomes.

Learn More:

brainly.com/question/11655383

brainly.com/question/13276299

brainly.com/question/13551177

brainly.com/question/11515691

#LearnWithBrainly

7 0

3 years ago

Read 2 more answers

In a unique species of plants, flowers may be yellow, blue, red, or mauve. All colors may be true-breeding. If plants with blue

Zolol [24]

Answer:

F1) 100% bbRr, red flowered plants.

Explanation:

<u>Available data:</u>

flowers may be yellow, blue, red, or mauve
colors may be true-breeding
the cross of blue-flowered plants with red-flowered plants, produce plants that have yellow flowers
F2 generation: 9/16 yellow, 3/16 blue, 3/16 red, and 1/16 mauve.

Knowing that the phenotypic ratio is 9:3:3:1, we can assume that there are two genes involved in the flower color expression. We can name these genes B and R with a dominant and a recessive allele each (B, b and R, r respectively).

According to the first cross, we might establish the following genotypes:

1st Cross: blue-flowered plant x red-flowered plant

Parentals) BBrr x bbRR

Gametes) Br, Br, Br, Br bR, bR, bR, bR

F1) 100% BbRr, yellow plants

Parentals) BbRr x BbRr

Gametes) BR, Br, bR, br

BR, Br, bR, br

Punnett square) BR Br bR br

BR BBRR BBRr BbRR BbRr

Br BBRr BBrr BbRr Bbrr

bR BbRR BbRr bbRR bbRr

br BbRr Bbrr bbRr bbrr

F2) 9/16 yellow ---> 1/16 BBRR + 2/16 BBRr + 4/16 BbRr + 2/16 BbRR

3/16 blue ------> 1/16 BBrr + 2/16 Bbrr

3/16 red---------> 1/16 bbRR + 2/16 bbRr

1/16 mauve ----> 1/16 bbrr

So,

Yellow-flowered plants: BBRR, BBRr, BbRR, BbRr
Red-flowered plants: bbRR, bbRr
Blue-flowered plants: BBrr, Bbrr
Mauve-flowered plants: bbrr

According to these genotypes, the second cross would be like following,

2nd Cross: true-breeding red-flowered plants with true-breeding mauve flowered plants.

Parentals) bbRR x bbrr

Phenotype) Red Mauve

Gametes) bR, bR, bR, bR br, br, br, br

Punnett square) bR bR bR bR

br bbRr bbRr bbRr bbRr

F1) 100% bbRr, red flowered plants.

6 0

3 years ago

In 1980, Y. published an extended pedigree of family in which

nydimaria [60]

Answer:

Zero (0)

Explanation:

According to the given information the genotype of the woman with blood type "AB" would by I^AI^B. The genotype of the man with blood type O would be "ii". Here, the alleles I^A and I^B are dominant over the allele "i".

A cross between parents with genotype I^AI^B and ii would give 50% of children with I^Ai genotype and 50% of children with I^Bi genotype. The children with "I^Ai genotype" would have blood type "A" and the children with I^Bi genotype would have blood type "B". This couple is never likely to have any child with blood type "O" since the mother does not carry allele "i".

Cross: I^AI^B x ii = 1/2 I^Ai : 1/2 I^Bi

5 0

3 years ago