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2 years ago

Describe the structure and the function of the stomach. Be sure to include cells and tissues relating to this organ.

Biology

1 answer:

Vikentia [17]2 years ago

8 0

Connective tissues support the tissues of the mucosa and connect it to the muscularis layer.

The blood supply of the submucosa provides nutrients to the wall of the stomach. Nervous tissue in the submucosa monitors the contents of the stomach and controls smooth muscle contraction and secretion of digestive substances.

<h3>What is the function of the stomach?</h3>

The stomach secretes acid and enzymes that digest food. Ridges of muscle tissue called rugae line the stomach.

The stomach muscles contract periodically, churning food to enhance digestion. The pyloric sphincter is a muscular valve that opens to allow food to pass from the stomach to the small intestine.

Thus, this could be the answer.

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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 two examples show adaptations for plants growing in the desert?

aleksklad [387]

there is no clear evidence about plants evolving in the desert, or fossils of any kind of cactus despite their distribution over a large area. But, scientists are of the opinion that they acquired specific attributes millions of years ago, as a response to the vastly changing climate and conditions.

-Desert plant root system

A typical desert plant has a deep root system, which is a specific physical mechanism. As the roots grow deep down the soil, they absorb soil water, which is then transported to the upper portions of the plant.

-Desert plant leaves

Bear small foliage with waxy surface to minimize transpiration (loss of water through stomatal pores), which in turn, helps to conserve water.

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3 years ago

_____ recognized the vital role of the internal environment and suggested that the objective of mechanisms within the body is to

VikaD [51]

Carolus Linnaeus recognized the vital role of the internal environment and suggested that objective of mechanisms within the body is to preserve the constant conditions of the internal environment.
He was later known as the creator of current modern system of naming animals called binomial nomenclature

hope this helps

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3 years ago

La maestra Leticia dice que para vivir en una democracia es indispensable saber lógica, por que requerimos saber argumentar para

Flura [38]

Translation??? Ingles.

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2 years ago

Explain how invertebrates can be used to monitor water pollution?

Ira Lisetskai [31]

Explanation:

Many invertebrate animals cant survive in water that has been polluted and resulted in a lowering of oxygen concentration so the presence or absence indicates the extent to which a body of water is polluted.

Lichens can be used to indicate air pollution as they grow exposed to places such us rocks and tree barks. They need to be efficient at absorbing water and nutrients to grow there. Rain water contains enough nutrients to keep them alive. Air pollutants dissolve in rain water especially sulfur dioxide can damage lichens and prevent them from growing. This makes them a natural indicator of air pollution.

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3 years ago

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