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vredina [299]
2 years ago
15

A black-eyed cat mates with an orange-eyed cat and produce all offspring with black eyes. Assuming these genes follow mendelian

genetics, what is the expected phenotypic ratio of a cross between two orange-eyed cats?
Biology
1 answer:
stiks02 [169]2 years ago
6 0

Answer:

all orange eyed

Explanation:

Mendelian genetics operate on the laws of segregation and independent assortment. This means that genes  from either parent segregates independently and are separated into gametes at an equal frequency.

Since a cross between all the black-eyed cats and orange eyed cats lead to all black eyed cats. It is safe to assume that the black eye phenotype is dominant and the orange eye phenotype is recessive. If two orange eyed cat crossed then all the offspring would be orange eyed.

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The brain volumes ​(cm cubed​) of 20 brains have a mean of 1116.2 cm cubed and a standard deviation of 127.7 cm cubed. Use the g
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Answer:

Yes

Explanation:

Range rule of thumb predicts the Range to be  a multiple of 4 of the standard deviation or to be four times  the standard deviation. Making the usual values equal to 2 standard deviations distanct of the mean of the data distribution.

In a given distribution with mean and standard deviation that is obtained, the usual values in mean (as seen in the attached image).

2*standard deviation and mean + 2*standard deviation.

If the data point is not up to the mean

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If the data point is more than the mean

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Let's take the xbar to be the mean and s as standard deviaiton

Given,

mean, xbar = 1116.2

standard deviation, s =127.7

The range rule of thumb shows that the usual values are within 2 standard deviations from the mean

Lower boundary

      = xbar - 2s

       = 1116.2 - 2(127.7)

       = 860.8

Upper boundary

      = xbar + 2s

       = 1116.2 + 2(127.7)

       = 1371.6

We should note that 1411.6 is not between 860.8 and 1371.6, which connotes that 1411.6cm^3 is unusually high.

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