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Nady [450]
4 years ago
10

What are the differences between aggregate fruit and multiple fruit ​

Biology
1 answer:
Dmitry_Shevchenko [17]4 years ago
4 0

Answer:

Aggregate fruit is a fruit that develops from several ovaries forming one flower. An example of an aggregate fruit is a raspberry. Multiple fruit is fruit that develops from a cluster of flowers, and each flower produces a fruit, which turns into a mass of a singular fruit. An example of multiple fruit is a pineapple. Each "cube" of a pineapple is a separate fruit formed from a single flower, yet the pineapple as a whole is one mass of pineapple.

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why is microwaved popcorn bad for pregnant women? please also talk about the effects it has on prenatal development.​
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Answer:

When exposed to high temperatures, fluorotelomers break down into PFOA(Perfluorooctanoic acid), a compound transmitted directly to the fetus which has been linked to developmental problems, as well as cancer, liver disease, and thyroid disease

Explanation:

5 0
3 years ago
What will happen if a drought causes the population of grass to decrease?
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Answer:

The other populations will decrease.

Explanation:

If the grass population decreases, then the grasshoper population will decreace, causing the mice population to decrease, which will lead to the decrease in owl population.

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What do you do if a hurricane is predicted?
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Answer:

You will need to prepare just incase it happens. Meaning buy grocceries, shelter and comfort items. And make sure to stay indoors at all times and stay away from windows or anything dangerous.

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3 years ago
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You are following segregation of five unlinked genes in a cross – A, B, C, D and E. Dominant alleles are capital letters and rec
bija089 [108]

Answer:

1/64 = 0.015625

Explanation:

Genotype of parents:  aa bb Cc Dd Ee x aa Bb Cc dd Ee

To calculate the frequency of the child with desired genotype, individual frequency of desired genotype can be calculated followed by application of product rule.

aa x aa = All aa

bb x Bb= 1/2 bb: 1/2 Bb

Cc x Cc= 1/4 CC: 1/2 Cc : 1/4 cc

Dd x dd = 1/2 Dd: 1/2 dd

Ee x Ee= 1/4 EE: 1/2 Ee: 1/4 ee

Expected frequency of the child with desired genotype= 1 aa x 1/2 bb x 1/4 cc x 1/2 dd x 1/4 ee = 1/2 x 1/4 x 1/2 x 1/4 = 1/64 = 0.015625

6 0
3 years ago
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