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2 years ago

What is the difference between Myeloid and Lymphoid hemopoiesis?

1 answer:

6 0

If the abnormal cells are primarily granulocytes (which destroy bacteria) or monocytes (which produce large cells called macrophages, which digest foreign substances and diseased cells), the leukemia is classified as myelogenous. If the abnormal cells are mainly lymphocytes (which orchestrate the immune system’s response to infection or disease), the leukemia is classified as lymphocytic.Leukemia arises from malfunctions in stem cells within the bone marrow that cause abnormal white blood cells to flood into the bloodstream. Leukemias are classified as either myelogenous (also called myeloid) or lymphocytic depending on which types of white blood cells are affected.

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What is an example of active transport

lubasha [3.4K]

Answer:

example of active transport is the sodium-potassium pump, which moves sodium ions to the outside of the cell and potassium ions to the inside of the cell.

3 0

2 years ago

Which of the following do ALL organisms have in commom

aliina [53]

Answer:

cells?

Explanation:

it would help if you gave the options :)

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2 years ago

Which of the following bacteria can grow in low oxygen environment: Salmonella Yersinia Listria monocytogenes Clostridium botuli

AlexFokin [52]

Answer: Clostridium botulinium

Explanation:

An anaerobic bacteria is an organism which does not require oxygen for it's survival. It may not survive or die in the presence of oxygen. These bacteria are used for the process of fermentation for the preparation of beverage and dairy products.

Clostridium botulinium is the example of the anaerobic bacteria. It is found in soil or marine environment in the form of spores. This bacteria is rod shaped. It forms the spores when the conditions for survival is poor. It causes a disease called as botulism.

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2 years ago

How does water clarity indirect effects on other species in food chain?

Vladimir [108]

Answer: The effects of which are devastating to human life, flora and fauna, which depend on these resources for their survival. Contamination of water causes disruptions of food chains and destruction of ecosystems. A deadly by-product of this water pollution is the spread of water-borne diseases such as cholera, typhoid etc.

Explanation:

3 0

1 year 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

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

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 these genes are linked.

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

2 years ago