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expeople1 [14]
3 years ago
14

7. How doessalt prevent food spoilage ?i will mark as brainlist

Biology
2 answers:
Semenov [28]3 years ago
8 0

Answer:

the salt dehydrate the food

PolarNik [594]3 years ago
3 0
Salt dehydrates the food therefore prevents food spoilage
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Hormone production and release are primarily controlled by negative feedback. In negative feedback systems, a stimulus elicits the release of a substance; once the substance reaches a certain level, it sends a signal that stops further release of the substance. In this way, the concentration of hormones in blood is maintained within a narrow range. For example, the anterior pituitary signals the thyroid to release thyroid hormones. Increasing levels of these hormones in the blood then give feedback to the hypothalamus and anterior pituitary to inhibit further signaling to the thyroid gland, as illustrated in Figure 18.14. There are three mechanisms by which endocrine glands are stimulated to synthesize and release hormones: humoral stimuli, hormonal stimuli, and neural stimuli.

Explanation:

Hyperthyroidism is a condition in which the thyroid gland is overactive. Hypothyroidism is a condition in which the thyroid gland is underactive. Which of the conditions are the following two patients most likely to have?

Patient A has symptoms including weight gain, cold sensitivity, low heart rate and fatigue.

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The term “humoral” is derived from the term “humor,” which refers to bodily fluids such as blood. A humoral stimulus refers to the control of hormone release in response to changes in extracellular fluids such as blood or the ion concentration in the blood. For example, a rise in blood glucose levels triggers the pancreatic release of insulin. Insulin causes blood glucose levels to drop, which signals the pancreas to stop producing insulin in a negative feedback loop.

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Suppose that one of the normal children from Part B and one of the albino children from Part C become the parents of eight offsp
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Full question from alternative source:

Albinism in humans is inherited as a simple recessive trait. for the following families, determine the most likely genotypes of the parents

Part A

Two normal parents have five children, four normal and one albino. What are the most likely genotypes of the parents?

Part B

A normal male and an albino female have six children, all normal. What is the most likely genotypes of the parents?

Part C

A normal male and an albino female have six children, three normal and three albino. What are the most likely genotypes of the parents?

Part D

Suppose that one of the normal children from Part B and one of the albino children from Part C become the parents of 8 offspring. What would be the expected number of normal and albino offspring?

Answer:

Part A - both heterozygous for the trait (Aa)

Part B - female aa, male likely AA

Part C - female aa, male Aa

Part D - 4 normal, 4 albino (50:50)

Explanation:

Part A

We are told that albinism is a recessive trait. Lets call it a and the normal allele A. Since neither of the parents have it, that must mean they each possess at least one copy of the dominant allele (either AA or Aa). Most of their children are normal, but 1 is albino. The albino child must have inherited two copies of the a allele, one from each parent. This means each parent must possess an a allele and must both be Aa.

Part B

Since the trait is recessive, we already know that the albino female must have the genotype aa. If her husband in homozygous normal (AA), all their children will be heterozygous (Aa), and therefore normal, which is the case in this cross. Therefore aa and AA are the most likely genotypes. However, it is still possible that the male has the genotype Aa. This would mean they had a 50:50 chance of producing an albino offspring. We know they have six normal children, so this is unlikely.

Part C

Since the trait is recessive, we know the female has the genotype aa. Unlike the last family, this family has some albino children. This means that the husband must be a carrier of the trait, but he is unaffected, so must be Aa.

As described in part B, if he is Aa and she is aa, they will have 50:50 normal:albino children. This is the ratio in this family.

Part D

The normal children from Part B are all heterozygous for the trait (Aa), inheriting one normal allele from their father, and one albino allele from their mother. The albino children from part C are all aa. Therefore, the cross is Aa x aa. We can draw a punnett square:

                  Aa

             A         a

       a   Aa       aa

aa   a   Aa       aa

Half the children have the genotype Aa, and half have the genotype aa. That means half will be normal and half will be albino. Therefore, 4 normal, 4 albino.

4 0
3 years ago
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