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Papessa [141]
4 years ago
15

Which row (a,b,c or d) in the table accurately categorizes the plant as gymnosperms and angiosperms

Biology
2 answers:
evablogger [386]4 years ago
8 0

Answer:

D

Explanation:

pishuonlain [190]4 years ago
7 0
Answer is "D".

Gymnosperms  are the plants which  produce naked seeds (unenclosed seeds). Examples are conifers, cycads, Ginkgo and gnetophytes. 

Angiosperms are flowering plants  which produce enclosed seeds. The seeds are covered by the fruit. Grains, flowering plants and fruit trees can be taken as the examples.
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Describe how cellular respiration and photosynthesis are related to one another.
Effectus [21]

Answer:

Photosynthesis makes the glucose that is used in cellular respiration to make ATP. ... While photosynthesis requires carbon dioxide and releases oxygen, cellular respiration requires oxygen and releases carbon dioxide. It is the released oxygen that is used by us and most other organisms for cellular respiration

8 0
3 years ago
Read 2 more answers
What is the scientific definition of energy? A. the ability to use the stored potential of an object B. the ability to use the c
dusya [7]
The answer is C. The ability to use an applied force in order to make an object move.
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3 years ago
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Assume a population of purple and white flower pea plants (2000 total plants) is in Hardy-Weinberg equilibrium. The frequency of
lys-0071 [83]

Answer:

a) 0.48

b)

p= 0.4\\q=0.6

c) 720

Explanation:

Given ,

Frequency of white flower plant (p^2) is 0.16

Thus, frequency of dominant white allele (p) is equal to \sqrt{0.16} \\= 0.4

As we that, as per  Hardy-Weinberg equilibrium equation

p+q=1

Substituting the value of p in above equation, we get

0.4+q=1\\q=1-0.4\\q=0.6

a) As per Hardy-Weinberg equilibrium equation

p^2+q^2+2pq=1\\

Substituting the given and calculated values in above equation, we get -

0.4^2+0.6^2+2pq=1\\0.16+0.36+2pq=1\\2pq=1-0.16-0.36\\2pq=0.48

b) The frequency (decimal form) of the dominant allele

p= 0.4\\q=0.6

c) number of homozygous purple flower plants

= frequency of purple flower plants (q^2) * Number of flowers

= q^2 * 2000\\= (0.6^2)*2000\\= 0.36 * 2000\\= 720

8 0
3 years ago
9/16 dark blue: 6/16 light blue: 1/16 white
Daniel [21]
IF THIS IS THE COMPLETE QUESTION WITH MULTIPLE CHOICE.

Two genes interact to produce various phenotypic ratios among F2 progeny of a dihybrid cross. Design a different pathway explaining each of the F2 ratios below, using hypothetical genes R and T and assuming that the dominant allele at each locus catalyzes a different reaction or performs an action leading to pigment production. The recessive allele at each locus is null (loss-of-function). Begin each pathway with a colorless precursor that produces a white or albino phenotype if it is unmodified. The ratios are for F2 progeny produced by crossing wild-type F1 organisms with the genotype RrTt.

9/16 Dark Blue 6/16 light blue 1/16 white

A) At least one copy of each dominant allele results in dark blue, at least one copy of either dominant allele produces light blue, and the absence of both dominant allele produces white.

B) If both dominant alleles are present, the result is dark blue. At least one copy of one specific dominant allele is required for light blue. If that dominant allele is not present, the result is white, regardless of whether the other dominant allele is present.

C) At least one copy of both dominant alleles results in dark blue; at least one copy of one of the dominant alleles also results in dark blue, but at least one copy of the other dominant allele produces light blue; and the absence of either dominant allele produces white.

MY ANSWER IS:
A) At least one copy of each dominant allele results in dark blue, at least one copy of either dominant allele produces light blue, and the absence of both dominant allele produces white.
3 0
4 years ago
Question 4
Dvinal [7]

Answer:

Oxygen

Explanation:

4 0
3 years ago
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