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guajiro [1.7K]
3 years ago
15

How would retaining extra cytoplasm enhance survival of a developing oocyte?

Biology
1 answer:
beks73 [17]3 years ago
6 0

Answer:  An extra cytoplasm enhance survival of a developing oocyte in such a way because cytoplasm serves as a reservoir of lipids, polysaccharide and lipids for the developing embryo.The more the cytoplasm content, more will be the nutrients for the developing embryo. The cytoplasm forms the yolk and provides nourishment to the developing embryo. As we know that the developing embryo requires more nutrition and the main source of nutrition is cytoplasm because the sperm has no reservoir of its own to provide nutrition to embryo.

Thus, an extra cytoplasm enhance the rate of survival of a developing oocyte.




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D. Water light and CO2 are needed for the process of photosynthesis.
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How does a photovoltaic cell convert light to electricity?
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I'm glad you asked!

Explanation:

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1. Flower position, stem length, and seed shape are three characters that Mendel studied. Each is controlled by an independently
maria [59]

Answer:

a. 1/64 AATTDD

b. 1/64 aattdd

c. 8/64=1/8 AaTtDd

d. 2/64 = 1/32 AATTDd or aattDd or AAttDd or aaTTDd

Explanation:

<u>Available data:</u>

  • Flower position: Axial (A) is dominant to Terminal (a)
  • Stem length: Tall (T) is dominant to dwarf (t)
  • Seed shape: Round (R) is dominant to wrinkled (r)

<u>Cross</u>:

Parental) AaTtDd   x   AaTtDd

Knowing that these four genes assort independently, then we can infer that the F1 will have the next genotypic proportions for each gene:

  • AA= 1/4

       Aa =2/4

       aa = 1/4

  • TT = 1/4

        Tt = 2/4

         tt = 1/4

  • DD = 1/4

        Dd = 2/4

        dd = 1/4

<em>This is the same as developing a Punnett square for each gene separately.</em>

Knowing this, to get each of the mentioned genotypes for each trait, we must multiply their respective genotypic proportions, like this:

a. homozygous for the three dominant traits, AATTDD

   AA1/4 x TT 1/4 x DD 1/4 = AATTDD 1/64

b. homozygous for the three recessive traits, aattdd

   aa 1/4 x tt 1/4 x dd 1/4 = aattdd 1/64

c. heterozygous for all three characters, AaTtDd

   Aa 2/4 x Tt 2/4 x Dd 2/4 = AaTtDd 8/16=1/8

d. homozygous for axial and tall, heterozygous for seed shape. There are four possibilities, all of them with the same result:

AATTDd --> 1/4  x 1/4 x 2/4 = 2/64=1/32

aattDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

AAttDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

aaTTDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

4 0
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Read 2 more answers
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sergejj [24]

Answer:

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Explanation:

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