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Fiesta28 [93]
3 years ago
7

A pea plant that has round seeds has the genotype Rr. It is crossed with a pea plant that has wrinkled seeds and the genotype rr

. What is the probability that the offspring will have wrinkled seeds?

Biology
1 answer:
o-na [289]3 years ago
7 0
Answer: The probability that the offspring will have wrinklied seeds is 50%.

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2. Describe the property of water that is indicated
Aleonysh [2.5K]

Answer:

The correct answer is - high specific heat.

Explanation:

Water atoms are polar, so they structure hydrogen bonds. This gives water remarkable properties, for example, a moderately high specific heat, density, adhesion, and cohesion.  

A water particle comprises of two hydrogen iotas clung to an oxygen molecule, and its general structure is bowed. Oxygen even bonding with hydrogen have two sets of unshared electrons. The entirety of the electron pair and unpaired repulse one another.  

Water's high specific heat is a property brought about by hydrogen holding among water particles. At the point when warmth is assimilated, hydrogen bonds are broken and water atoms can move uninhibitedly. At the point when the temperature of water diminishes, the hydrogen bonds are shaped and discharge a lot of vitality.

Thus, the correct answer is - high specific heat.

5 0
3 years ago
Small hydrophilic molecules are transported through a cell membrane by
o-na [289]
Proteins which form channels may be utilized to enable the transport of water and other hydrophilic molecules; these channels are often gated to regulate transport rate. In facilitated transport, hydrophilic molecules bind to a "carrier" protein; this is a form of passive transport
4 0
3 years ago
How long would it take
Valentin [98]
Solution:
4.2 x 10^6 bp/10^3 bp/seconds         = 4.2 + 103 s which is 4200 seconds and equivalents to 70 minutes

In addition, assuming a pause of 2 seconds for re initiating after completing every okazaki fragment and assuming the okazaki fragments average 1000 nucleotide long.  
4.2 x 10^6 bp/10^3 bp                        = 4200 okazaki fragments                           4200 * 2 seconds                               = 8400 seconds which is 140 minutes or 2 hours 20 minutes of pauses alone.                                                                        

Therefore, overall time would be pauses plus the 70 minutes so total time of 210 minutes. Assuming that the replisome completely disassociates from the DNA after every okazaki fragment and must spend one-minute rebinding.

4200 okazaki fragments. 60 seconds rebinding time per fragment:                        4200 x 1 minute = 4200 minutes rebinding time plus 70 minutes for actual replication. 4200 minutes is 70 hours which is almost 3 days. 
3 0
3 years ago
(25pts) I will give the BRAINLEST
Mashcka [7]

Answer: (D) Only cell B forms a cell plate during cytokinesis.

Explanation: The “cell cycle” describes the process that cells go through, from their “birth” as new daughter cells, until they themselves are ready to split and become “parent cells” to two new daughter cells. The formation of the cell plate takes place during the mitotic phase.

8 0
3 years ago
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