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grandymaker [24]
3 years ago
15

Here’s part b of what I asked before

Biology
1 answer:
dedylja [7]3 years ago
6 0

Explanation:

All cells have a plasma membrane, ribosomes,and DNA. The plasma membrane, or cell membrane, is the phospholipid layer that surrounds the cell and protects it from the outside environment. They both use ribosomes for producing proteins.

+sorry i made a mistake on the previous question. Prokaryotic cell doesn't have nucleus. I obviously didn't know the right meaning of this word bc im not a native english speaker

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Juliette [100K]

Answer:

4 the cube with many circles in it or box C

3 objects that float paperclip, feather, air molecules, sticker  objects that sink book, phone, human, plate.

2 ?

Explanation:

4 0
2 years ago
The ABO blood group assorts independently of the Rhesus (Rh) blood group and the MN blood group. Three alleles, IA , IB , and i,
Eva8 [605]

Answer:

Part 1:

The mother’s genotype is ii, rr, MN

The father’s genotype is IAIA/IAi, Rr, MM

Part 2:

Approximately 1/32 (3.125%) of children can be expected to have the stated blood types.

Explanation:

Part 1:

The mother has blood types 0, Rh- and MN. Therefore, her genotype has to be ii, rr, MN.

The father has ABO type A. His ABO genotype may therefore be either IAIA or IAi, as both are compatible with type A offspring with a type 0 mother. His Rh genotype must be rr, since his Rh- child necessarily has to have inherited one r allele from each parent. Finally, his MN genotype has to be MM since both alleles are codominant.

Part 2:

Using the given data, we build the following Punnett chart:

                IA R N     IA r N     i R N     i r N    

IA R M

IA R N

IA r M

IA r N

IB R M

IB R N

IB r M                                                    X

IB r N                                                              

Note that, of all 32 possible combinations, only a single one of them (marked with “X” produces a genotype which matches the desired blood type, as follows:

genotype: IBi, rr, MN

blood type: B, Rh-, MN

3 0
3 years ago
A bacterial sample has a cell density of 1.85 x106 CFU/mL. What plate dilution should yield a countable plate? Which two dilutio
insens350 [35]

Answer:

Explanation:

Bacterial count in stock- 1.85x10^6 cfu/ml

Dilution methods

Take 100 uL or (0.1ml) from stock and add to 900ul (0.9ml) saline and mixed it- this makes 10^1dilution.

Now take 100ul from 10^1 dilution and add to next 900ul saline this is 10^2 dilution, similarly do upto 10^5 dilution.

Then take 100ul from 10^ 4 and 10^5 dilution seperately and plate on LB agar plate seperetely and count the colonies.

Cfu/ml formula= (No.of colonies x dilution factor)/0.1 ml

So suppose, 18 colonies formed on 10^4 dilution then total no. Of cells in stock will be 18x10^4/ 0.1= 18x10^5 cfu/ml.

If we dilute 10^4 or 10^5 that's leads to colony count of 18-19 colonies on 10^4 dilution while 2 colonies should come on plate of 10^5 dilution.

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3 years ago
Which is required for both anaerobic respiration and aerobic respiration?
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.........................................................ATP
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