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3 years ago

HELP ASAP!!!!!!

Biology

2 answers:

Eva8 [605]3 years ago

7 0

Answer:

C-Attaching to a host

Explanation:

It wouldn’t be entering the cell because I’ve learned that it needs to attach to a host first, and it wouldn’t be replicating because it needs to be in the host before it starts. Which leaves assembling and attaching. It is attaching and not assembling because assembling is one of the last ones, second to last step, so the answer would be attaching.

stepladder [879]3 years ago

4 0

Answer:

The answer is D I am pretty sure plz help me with the question

Explanation:

A tomato plant grew 6 inches the first week, before slowing to an average of 0.5 inch per day. How many days will it take before the tomato plant reaches a height of 16 inches? Use the equation 16 = 0.5d + 6

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The Kingdom Fungi is found under which domain?

ValentinkaMS [17]

Answer:

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Do you think mouse offspring will always look like their parents? ______________________

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3 years ago

Considering the same population of cats as in Part A, what is the expected frequency of each genotype (TLTL, TLTS, TSTS ) based

zaharov [31]

Answer:

P = f(TLTL) = 0,16

H = f(TLTS) = 0,48

Q = f(TSTS) = 0,36

Explanation:

Hello!

The allele proportion of any locus defines the genetic constitution of a population. Its sum is 1 and its values can vary between 0 (absent allele) and 1 (fixed allele).

The calculation of allelic frequencies of a population is made taking into account that homozygotes have two identical alleles and heterozygotes have two different alleles.

In this case, let's say:

f(TL) = p

f(TS) = q

p + q = 1

Considering the genotypes TLTL, TLTS, TSTS, and the allele frequencies:

TL= 0,4

TS= 0,6

Genotypic frequency is the relative proportion of genotypes in a population for the locus in question, that is, the number of times the genotype appears in a population.

P = f(TLTL)

H = f(TLTS)

Q = f(TSTS)

Also P + H + Q = 1

And using the equation for Hardy-Weinberg equilibrium, the genotypic frequencies of equilibrium are given by the development of the binomial:

$p^{2} = f(TLTL)$

$2pq = f(TSTL)$

$q^{2} = f(TSTS)$

So, if the population is in balance:

$P = p^{2}$

$H = 2pq$

$Q = q^{2}$

Replacing the given values of allele frecuencies in each equiation you can calculate the expected frequency of each genotype for the next generation as:

$f(TLTL) = P = p^{2} = 0,4^{2} = 0,16$

$f(TLTS) = H = 2pq = 2*0,4*0,6 = 0,48$

$f(TSTS) = Q = q^{2} = 0,6^{2} = 0,36$

I hope you have a SUPER day!

6 0

3 years ago