A gland is a group of cells or an organ that

What are some similarities of sponges and cnidarians

Svetlanka [38]

Similarities:

They contain nematocyst (cells that sting)

some are sessil/free swimming

Some corals have a symbiotic relationship with algae, they need to grow in shallow waters so that light can penetrate for the algae to photosynthesize.

5 0

2 years ago

Which of the following is/are common to chemiosmosis and the light-dependent reactions of photosynthesis? electron transport onl

blondinia [14]

Answer:

Both electron transport and a proton gradient

Explanation:

The process of oxidative phosphorylation in mitochondria and electron transport chain in photosynthesis undergo chemiosmosis to produce ATP molecules.

Chemiosmosis is a process where the energy utilized by the movement of proton and electrons produces ATP molecules.

Both the processes involve the movement of electrons through electron carriers where the reduced energy is utilized to drive the flow of protons through the plasma membrane. This creates a proton gradient across the plasma membrane which rotates the ATP synthase and converts the ADP molecules into ATP molecules.

Thus, the selected option is correct.

3 0

2 years ago

Which wold tell you for sure if you had a sample of real gold

Advocard [28]

To perform this test to check if gold is real, make a light, tiny scratch on the item using a small nail file. Choose a spot that is not noticeable if possible. Use a dropper to apply a small drop of nitric acid to the scratch. If there is no reaction, then the piece is probably made from real gold.

5 0

2 years ago

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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:

Available data:

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)

Cross:

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

This is the same as developing a Punnett square for each gene separately.

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

2 years ago

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HELP

anastassius [24]

Answer:

the answer is d

Explanation:

3 0

2 years ago

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