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Red flowers (r) and white flower (w) are crossed to produce a pink flower (rw). what is the probability of a red flower, if 2 pi

Dafna11 [192]

I think the probability will be 0 then.

7 0

2 years ago

When you were testing the ciliospinal reflex and only stimulated the left side of the subject, was the response reflex contralat

natulia [17]

Options: A. contralateral

B. ipsilateral

C. both

Answer:B. ipsilateral

Explanation:The ciliospinal reflex which is also known bas pupillary-skin reflex is usually made uo of the dilation of the ipsilateral pupil this dilation is in response to pain applied to the neck, face, and upper trunk.

If you stimulate the left side of the subject, the response will be only in the ipsilateral which means the stimulation will be felt in the same side of the subject which in this case is the Left side if the subject.

5 0

3 years ago

What is the most likely evidence of an energy transformation taking place when a wire that is twisted and turned breaks?

Mnenie [13.5K]

The answer is C

Hope this helps!

3 0

3 years ago

Read 2 more answers

3 The plasma membrane is semipermeable because -

Irina-Kira [14]

Answer: B) proteins in the membrane act like gates to control entrance and exit of molecules.

Explanation: Plasma membrane is composed of phospholipids bilayer in which the hydrophobic regions of the phospholipids face each other to form the core of the bilayer. Proteins are embedded in the bilayer through an interaction between the amino acid side chains and the nonpolar acyl chains of the phospholipids. The embedded membrane proteins form channels for passage of ions and other molecules. Each membrane protein is selected for one or more ions or molecules that passes through it. The membrane is said to be semipermeable because it allows the passage of some molecules and excludes the passage of others.

6 0

3 years ago

Suppose that a strain of bacteria lacks a key biochemical used in purine synthesis. Complete all the statements about purine syn

olga55 [171]

Answer:

1. Phosphoribosyl amine

2. glycinamide ribonucleotide (GAR), a formyl group

3. Formylglycinamide ribonucleotide (FGAR), an amino group

4. Carboxyaminoimidazole ribonucleotide (CAIR), an amino group.

Explanation:

1. Phosphoribosyl amine will accumulate in bacteria that lack glycine.

2. The intermediate glycinamide ribonucleotide (GAR) will build up in the absence of tetrahydrofolate.

N10‑Formyltetrahydrofolate donates a formyl group to the substrate

3. Formylglycinamide ribonucleotide (FGAR) will accumulate in the absence of glutamine.

Glutamine donates an amino group

4. The intermediate carboxyaminoimidazole ribonucleotide (CAIR) will accumulate when aspartate is lacking.

Aspartate donates an amino group.

8 0

3 years ago