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Trava [24]
3 years ago
6

The process indicated by the letter _____ produces a diploid structure. A life cycle of the black bread mold. Each letter marks

a definite process of this cycle. Letter A marks a process of spore germination. Letter B marks a process of cytoplasm fusion. Letter C marks fusion of nuclei and formation of a round solid structure. Letter D marks a process by which the round solid structure produces spores. Letter E marks the other cycle, by which mycelium is formed from the spores. The process indicated by the letter _____ produces a diploid structure. A life cycle of the black bread mold. Each letter marks a definite process of this cycle. Letter A marks a process of spore germination. Letter B marks a process of cytoplasm fusion. Letter C marks fusion of nuclei and formation of a round solid structure. Letter D marks a process by which the round solid structure produces spores. Letter E marks the other cycle, by which mycelium is formed from the spores. A B C D E
Biology
1 answer:
Gnoma [55]3 years ago
4 0

The right answer is the letter B.

Cytogamy, in animals and in lower plants, indicates the fusion of two gametic cells (n). Thus, their genetic material (the two nuclei) will be in the same cytoplasm, and so we will have a set of two chromosomes (2n) in a cell, and this is the definition of a diploid cell.

Then the two nuclei will merge (karyogamy) to give the zygote, which will eventually give a new species.

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bixtya [17]

Answer:

a maybe correct me if im wrong

6 0
3 years ago
A Mother who is homozygous for Huntington's disease and a father who is homozygous normal want to have a baby. What genotypes an
natita [175]

Answer: The possible genotype for their child is Hh. All the children will be heterozygous for Huntington (Hh).

The possible phenotype for their child is Huntington disease.

Explanation: If H represents the trait for Huntington and h represents normal trait; and if Huntington trait (H) is dominant over normal trait (h), therefore the genotype of the mother who is homozygous for Huntington disease is HH and the genotype of the man who is homozygous normal is hh.

A cross between the man and woman will produce offsprings who are all heterozygous for Huntington disease Hh. Phenotypically, the offsprings will manifest as Huntington disease.

See the attached punnet square for more information

6 0
3 years ago
One reason that mendel chose garden peas for his experiments on heredity was because
Leto [7]

A colorimetric method based on tetrazolium salts (e.g., MTT, XTT, CCK-8/WST-8) are especially useful for assaying the quantification of viable cells.

https://www.creative-bioarray.com/support/colorimetric-cell-viability-assay.htm

8 0
4 years ago
What stage of respiration immediately follows the process depicted below? Two pyruvate molecules undergo a chemical reaction, co
Nadya [2.5K]

Answer:

Electron transport chain

Explanation:

Cellular respiration is the process whereby glucose is broken down in the cell to yield energy in form of ATP. This process, which is important to the metabolic functioning of every organism, occurs in three major stages namely: glycolysis, kreb's cycle and electron transport chain (ETC).

According to this question, a stage of respiration is described as follows:

- Two pyruvate molecules undergo a chemical reaction, combining with coenzyme A to form two acetyl-CoA molecules.

- Two carbon dioxide molecules and two NADH molecules are formed as a result of this process.

The process depicted above is KREB'S CYCLE of cellular respiration, hence, the stage of respiration that immediately follows the process is called ELECTRON TRANSPORT CHAIN (ETC).

3 0
3 years ago
Read 2 more answers
Consider a cross between a plant with two flower colors, yellow and red. Yellow (Y) exhibits complete dominance over red (y). It
Aneli [31]

Answer:

1) option a is correct. (260+270)/1545

<em>(Note about this option: In the statement, it is written as </em><em>(260+270) 1545. </em><em>Lacks the division symbol, </em><em>/ </em><em>)</em>

2) From these results one can conclude that the two genes are linked (option a)

Explanation:

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

We might verify which are the recombinant gametes produced by the di-hybrid, and we will be able to recognize them by looking at the phenotypes with lower frequencies in the progeny. Phenotypes with the highest frequencies represent the parental gametes.

To calculate the recombination frequency, we will use the following formula: P = Recombinant number / Total of individuals.  

In the present example:

Parental)        YySs                        x                 yyss

Gametes)  YS parental type                        ys, ys, ys, ys

                 ys parental type

                 Ys recombinant type

                 yS recombinante type

The total number of individuals in the offspring: 1545

Phenotypic class Number of offspring  

  • 500 Y-S- (parental)
  • 515 yyss (parental)
  • 260 Y-ss (recombinant)
  • 270 yyS- (recombinant)

According to this information, the phenotypic frequencies of the progeny differ from the phenotypic ratio 1:1:1:1. We can assume then, that t<u>hese genes are linked. </u>

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 260 + 270 / 1545

P = 530/1545

P = 0.343

The genetic distance between genes is 0.343 x 100= 34.4 MU.

6 0
3 years ago
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