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3 years ago

When conducting a biology investigation why is it important to note the safety rules and symbols

Biology

2 answers:

bezimeni [28]3 years ago

5 0

Answer:

The safety rules and symbols have to be noticed when there is a biology investigation being carried on. Because by interpreting them risk can be minimized and unfortunate situations avoided.

Explanation:

To understand this we need to analyze it from a Prevention perspective. First of all, prevention is the best effort to reduce risks associated with a certain situation. In the case of a biological investigation, people have to know all it's meanings because they can know the risks. As well as to be prepared to react in case something happens. Because even though we are carrying out the best to prevent a misfortunate situation they can still happen.

Free_Kalibri [48]3 years ago

3 0

To avoid any injuries or emergencies

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What are 2 ways in which gametes differ from somatic cells?

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Second difference between them is that gametes are products of meiosis, while somatic cells are products from mitosis. Both are types of cell divisions but meiosis produces 4 genetically different cells while mitosis produces 2 genetically indentical daughter cells.

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3 years ago

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Answer:

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3 years ago

Which of the following actions would be most likely to increase the protection of ecological systems in national forests?

insens350 [35]

Answer:

The correct answer is - option 3.

Explanation:

The given options are the question are affects the protection of the ecological system in one or another manner but the mandating the rights of government sale of the logging rights in a national forest must be a proper and fair market value.

This will lead to a decrease in the illegal sale of logging sale which results in maintaining the ecological system in the national forest.

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3 years ago

Which statement about the starch debranching enzyme is correct?

Roman55 [17]

Answer:

B) If two non functional copies are inherited, the pea will be wrinkeld.

Explanation:

The gene for starch debranching is one of the genes that regulate more than one genetic trait. The starch debranching gene also regulates the seed shape in peas. The gene has two alleles. The allele B is completely dominant over allele "b" with respect to seed shape. Both homozygous and heterozygous dominant genotypes (BB and Bb) produce smooth seeds while homozygous recessive genotype (bb) imparts wrinkled shape to the seeds.

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2 years ago

The Offspring produced by a cross between two given types of plants can be any of the three genotypes denoted by A, B, and C. A

ratelena [41]

Complete question:

The offspring produced between two given types of plants can be any of the three genotypes, denoted by A, B and C. A theoretical model of gene inheritance suggests that the offspring of types A, B and C should be in a 1:2:1 ratio (meaning 25% A, 50% B, and 25% C). For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded in the table below.

Genotype Observed frequency

A → 18 individuals

B → 55 individuals

C → 27 individuals

Do these contradict the genetic model?

Use a 0.05 level of significance.

Determine the chi-square test statistic.

Answer:

Do these contradict the genetic model? No, according to the chi-square test, there is not enough evidence to reject the null hypothesis of the population being in equilibrium.

Explanation:

Available data:

Crossed genotypes: two
Genotypes among the offspring; Three → A, B, and C
Expected phenotypic ratio → 1:2:1
Total number of individuals, N = 100
A = 18 individuals
B = 55 individuals
C = 27 individuals

So, let us first state the hypothesis:

H₀= the population is equilibrium for this locus → F(A) = 25%, F(B) = 50%, F(C) = 25%
H₁ = the population is not in equilibrium

Now, let us calculate the number of expected individuals, according to their expected ratio.

4 -------------- 100% -------------100 individuals

1 --------------- 25% -------------X = 25 individuals A

2 -------------- 50% -------------X = 50 individuals B

1----------------- 25%--------------X = 25 individuals C

A B C

Observed 18 55 27
Expected 25 50 25
(Obs-Exp)²/Exp 1.96 0.5 0.16

(Obs-Exp)²/Exp

A) (18 - 25)²/25 = 49/25 = 1.96

B) (55 - 50)² / 50 = 25/50 = 0.5

C) (27 - 25)²/25 = 4/25 = 0.16

Chi square = X² = Σ(Obs-Exp)²/Exp

∑ is the sum of the terms
O are the Observed individuals: 2 in chamber B, and 18 in chamber A.
E are the Expected individuals: 10 in each chamber

X² = ∑ ((O-E)²/E) = 1.96 + 0.5 + 0.16 = 2.62

Freedom degrees = 2

Significance level, 5% = 0.05

Table value/Critical value = 5.99

X² < Critical value

2.62 < 5.99

These results suggest that there is not enough evidence to reject the null hypothesis. We can assume that the locus under study in this population is in equilibrium H-W.

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3 years ago