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3 years ago

6

What does it mean if you have a brown haired child

1 answer:

Verizon [17]3 years ago

8 0

Answer:

Nothing.

Explanation:

You just have a brown haired child, it means nothing.

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A fossil is discovered that has only 12.5% of the carbon-14 that it would have had originally (when the animal was alive). If th

mariarad [96]

Answer:

The fossil is 17,100 years old.

Explanation:

The decay equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt}= -\lambda N$

$\Rightarrow \frac{dN}{N}= -\lambda dt$

Integrating both sides

$\Rightarrow \int\frac{dN}{N}= \int-\lambda dt$

$\Rightarrow ln |N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$ln |N_0|=-\lambda .0+c$

$\Rightarrow ln |N_0|=c$

$\therefore ln |N|=-\lambda t+ln|N_0|$

$\Rightarrow ln |N|-ln|N_0|=-\lambda t$

$\Rightarrow ln |\frac {N}{N_0}|=-\lambda t$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow N=N_0e^{-\lambda t}$

The decay equation is

$N=N_0e^{-\lambda t}$

Given that,

The half life of carbon - 14 is 5700 years.

For half life, $N=\frac{1}{2} N_0$

To find the value of $\lambda$ , we need to put the value of N and t in the decay equation.

$\frac12N_0=N_0e^{-\lambda \times 5700}$

$\Rightarrow \frac12=e^{-\lambda \times 5700}$ [ Divided $N_0$ both sides]

Taking ln both sides

$\Rightarrow ln| \frac12|=ln|e^{-\lambda \times 5700}|$

$\Rightarrow ln| \frac12|={-\lambda \times 5700}$

$\Rightarrow \lambda= \frac{ln| \frac12|}{-5700}$

$\Rightarrow \lambda= \frac{ln|1|-ln|2|}{-5700}$ [ $ln|\frac mn|= ln |m|-ln |n|$ ]

$\Rightarrow \lambda= \frac{ln|2|}{5700}$ [ln 1= 0]

The fossil has only 12.5% of the carbon carbon-14 that it would have had originally.

So, $N=\frac{12.5}{100} N_0$

Then,

$\frac{12.5}{100} N_0=N_0e^{-\frac{ln|2|}{5700}t$

$\Rightarrow \frac{12.5}{100} =e^{-\frac{ln|2|}{5700}t$

Taking ln both sides

$\Rightarrow ln|\frac{12.5}{100} |=ln|e^{-\frac{ln|2|}{5700}t}|$

$\Rightarrow ln|\frac{12.5}{100} |={-\frac{ln|2|}{5700}t}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|} {-\frac{ln|2|}{5700}}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|\times 5700} {-{ln|2|}}$

$\Rightarrow t=17,100$

The fossil is 17,100 years old.

7 0

4 years ago

the idea that the sequence of bases in dna specifies the sequence of bases in an rna molecule, which specifies the sequence of a

Mkey [24]

The answer is the central dogma of molecular biology.

The central dogma of molecular biology explains that the sequence of DNA specifies the sequence of mRNA, which, further, specifies codes for the sequence of proteins. Simply, DNA codes for RNA and RNA codes for proteins. The central dogma was first stated by Crick in 1956.

7 0

3 years ago

These may all be classified as

Monica [59]

What is being classified???

6 0

3 years ago

Which of the following are examples of asexual reproduction?

konstantin123 [22]

The last 2 answers I believe and Srry if I’m not right

5 0

3 years ago

Read 2 more answers

. In this example, red flower coloring is dominant to white, and your population of 200 plants has 150 plants with red flowers.

AnnyKZ [126]

Answer:

a) The allele frequencies are - $p= 0.8660\\q= 0.134$

b) Genotype frequencies for this population- $p^2=0.75, q^2= 0.0179, 2pq = 0.2321$

Explanation:

Given ,

Red color of flower is dominant to white color flower

Let "R" represents allele for red color and "r" represents allele for white color

out of $200$ flower, $150$ are red flowers.

So the frequency of red flowers with genotype "RR" is equal to

$p^{2} = \frac{150}{200} \\p^{2} =0.75$

As per hardy Weinberg's I equilibrium equation -

$p+q=1\\p=\sqrt{p^2} \\p=\sqrt{0.75} \\p= 0.8660\\q= 1-p\\q= 1-0.8660\\q = 0.134$

a) The allele frequencies are - $p= 0.8660\\q= 0.134$

Frequency of genotype "rr" i.e homozygous recessive white flower is

$q^{2} \\0.134^2\\q^2 = 0.0179\\$

Substituting the values of $p^2$ and $q^2$ , in below equation, we get -

$p^2+2pq+q^2=1\\2pq= 1-p^2-q^2\\2pq= 1-0.75-0.0179\\2pq=0.2321$

b) Genotype frequencies for this population- $p^2=0.75, q^2= 0.0179, 2pq = 0.2321$

8 0

3 years ago