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boyakko [2]
3 years ago
12

how are tears a example of irreducible complexity? and why could it have not happened through evolutionary processes?

Biology
1 answer:
butalik [34]3 years ago
8 0
Here i hope this helps
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"To restore and maintain the chemical, physical, and biological integrity of the nation's water" was the purpose of
kompoz [17]

The answer is d..........

4 0
3 years ago
There are 252 deer in a population. There is no net immigration or emigration. If 47 deer die and 32 deer are born in one month,
morpeh [17]

Answer:

240

Explanation:

252-47=205

205+32=237

237 to the nearest whole number is 240

hope it helps

7 0
3 years ago
Calculate the final concentration of BSA in the problems below using the formula
Rudiy27

Answer:

A. C_2=1.5\frac{mg}{mL}

B. C_2=0.075\frac{mg}{mL}

C. C_2=0.01\frac{mg}{mL}

D. C_2=0.001\frac{mg}{mL}

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

C_2=\frac{C_1V_1}{V_2}

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}

Best regards.

7 0
3 years ago
Given the following cross TtYyRr x TtyyRr (T = tall; t = short Y = yellow; y = green R = round; r = wrinkled), what proportion o
mihalych1998 [28]

Answer:

3/32 ttyyR-  

Explanation:

Cross: Tall, Yellow, Rounded individuals with a tall, green, rounded individual

Parentals) TtYyRr      x      TtyyRr      

Gametes) TYR, TyR, TYr, Tyr, tYR, tyR, tYr, tyr (Parent one)

                 TyR, TyR, Tyr, Tyr, tyR, tyR, tyr, tyr (Parent two)

We need to know what proportion of offspring is expected to be short plants with round, green seeds. So we need to identify the gametes for these traits. The genotypes are:

  • Shot → tt
  • Round → RR or Rr
  • Green → yy

⇒ Parent one can provide gametes tyR and tyr

⇒ Parent two can provide gametes tyR and tyr

(1/8 tyR x 2/8 tyR) + (1/8 tyR x 2/8 tyr) + (1/8 tyr x 2/8 tyR) =

2/64 ttyyRR + 2/64 ttyyRr + 2/64 ttyyRr =

1/32 ttyyRR + 2/32 ttyyRr =

3/32 ttyyR-          

4 0
4 years ago
It takes the Earth
netineya [11]

Answer:

The answer is D. One year

8 0
3 years ago
Read 2 more answers
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