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3 years ago

Matching/Sequencing Activity: Maslow's Hlerarchy of Needs

Biology

1 answer:

gogolik [260]3 years ago

6 0

Answer:

Explanation:

1. Matching each label to corresponding level of Maslow's Hierarchy, we have

a. relationships with others and giving affection ------ Love and Belongingness

b. confidence in oneself ------- Self-esteem

c. risk avoidance and protection---Safety and security

d. self-acceptance and expression---- Self-actualization

e. food and shelter-----Physiological Needs

2.rearranging the levels so that they are in the correct order, with the lowest level of the Hierarchy on the bottom.

Maslow's hierarchy of needs is a psychology theory which shows and explains human needs in a five-tier model pyramid in hierarchy From the top to the bottom of the hierarchy, we have

1.self-acceptance and expression---- Self-actualization

2.confidence in oneself ------- Self-esteem

3.relationships with others and giving affection ------ Love and Belongingness

4.risk avoidance and protection---Safety and security

5. food and shelter-----Physiological Needs

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3 years ago

In Drosophila, vestigial (partially formed) wings (vg) are recessive to normal long wings (vg+) and the gene for this trait is a

Marina CMI [18]

Answer/ Explanation:

a. The genotype of a homozygous white eyed long winged female would be Vg+Vg+XrXr. We denote the white allele as recessive (r) because the XY male only has one copy and yet has red eyes, so the red eye trait (R) must be dominant. A homozygous red eyed vestigial winged male would have be VgVgXRY. The possible gametes for the female are Vg+Xr only. For the male, the possible gametes are VgXR or VgY

The attached punnett square shows the results of the cross. The females will all be Vg+VgXRXr. The males will all be Vg+VgXRY (must inherit Y from father). That means they will all have normal length wings, the males will have white eyes and the females will have red eyes.

b. The F2 flies arise from intercrossing the F1, so the cross will be Vg+VgXRXr x Vg+VgXRY. The possible gametes for the mother are: Vg+XR, Vg+Xr, VgXR or VgXr. The possible gametes for the father are Vg+Xr

, Vg+Y

, VgXr

, VgY

. The attached punnet square shows this cross. The ratio of the phenotypes will be 6:6:2:2, or 3:3:1:1 (long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye), genotypes shown in the attachment.

c. F1 cross back to the mother would be Vg+VgXRY x Vg+Vg+XrXr. The genotypes are shown in the attached punnet square. The offspring will all be long-winged with white eyes. The F1 to the father would be Vg+VgXRXr x VgVgXRY. The ratio would be 3:3:1:1 long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye

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3 years ago

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Alekssandra [29.7K]

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3 years ago

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