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LenKa [72]
2 years ago
11

On which colors is it equally likely that the spinner will land? red and blue green and red blue and green yellow and green

Biology
1 answer:
NNADVOKAT [17]2 years ago
5 0

There is an equal chance that the spinner will stop on either the red or blue segment.

<h3>On which colors is it equally likely that the spinner will land?</h3>

Probability theory is the mathematical study of the numerical representation of the probability that an event will take place or a statement will be true. Probability is measured on a scale from 0 to 1, with 0 denoting complete impossibility and 1 denoting absolute certainty.

Generally, the equation for the Probability of getting of  spinner landing is  mathematically given as

P= \frac{ total \ number \ of \ favourable \ outcome}{total \ number \ of  \ outcome }

Therefore

For P(Red)=3/12

P(Green)=4/12

P(blue)=3/12

P(yellow)=2/12

In conclusion, There is an equal chance that the spinner will stop on either the red or blue segment.

Read more about probability

brainly.com/question/795909

#SPJ1

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