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2 years ago

The plasmodesmata in plants are functionally most similar to which animal cell junction?

Biology

1 answer:

morpeh [17]2 years ago

5 0

Answer:

B.) Gap junction

Explanation:

Plasmodesmata (singular form: plasmodesma) are intercellular organelles found only in plant and algal cells. (The animal cell "equivalent" is called the gap junction.)

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At a chiasma, two ________ are attached to each other.

babymother [125]

Homologous or sister chromatids

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3 years ago

13. Heme is an achiral molecule that binds to the protein myoglobin. However, a 30 micromolar solution of myoglobin (containing

Anastaziya [24]

Answer:

The near-UV CD range (>250 nm) of proteins delivers info on the define configuration. The indications found within the 250–300 nm section are because of the engagement, dipole alignment and also the environment of the encircling atmosphere of the essential amino acid, tyrosine, aminoalkanoic acid (or S-S disulfide associations) and essential amino acids. In contrast to far-UV CD, the near-UV CD variation can't be allotted to any specific 3D configuration. Relatively, near-UV CD varieties offer essential info on the character of the prosthetic teams in proteins, e.g., the pigment teams in hemoprotein and cytochrome.

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3 years ago

If all the leaves are removed from a lycophyte, which process would affected?

Pepsi [2]

Respiration and photosynthesis because both happen through the leaves of any plant.

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3 years ago

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When the football is snapped, which player will have the smallest change in kinetic energy?

DanielleElmas [232]

When you are finding the kinetic energy, it is best to keep this formula in mind.

Ek = 1/2mv²

Where m is mass and v is velocity. Basically sub the numbers in to find the kinetic energy for each. In this case, subbing all the numbers, the smallest change in kinetic energy is the Lineman with a kinetic energy of just 540N.

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2 years ago

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22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit

mylen [45]

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square ( $X^2$ ):

$X^2$ = $\frac{(O - E)^2}{E}$ where O = observed frequency and E = expected frequency.

The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.

For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

phenotype (1) = 9/16 x 293 = 164.81

phenotype (2) = 3/16 x 293 = 54.94

phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The $X^2$ is calculated thus:

Phenotype O E $X^2$

1 88 164.81 $\frac{(88 - 164.81)^2}{164.81}$ = 35.80

2 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

3 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

4 81 18.31 $\frac{(81 - 18.31)^2}{18.31}$ = 214.64

Total $X^2$ = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

$X^2$ tabulated = 7.815

<em>The calculated </em> $X^2$ <em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently. </em>

4 0

2 years ago