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2 years ago

:3 what is the biggest cat in the world?

Biology

2 answers:

Elenna [48]2 years ago

7 0

Answer:

This cat right here

Pani-rosa [81]2 years ago

5 0

Yours :) :) :) okay that’s a weird question

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You discover a population of mustard plants and find it to be at Hardy–Weinberg equilibrium with respect to the R locus. Suppose

natulia [17]

Answer:

The fraction of heterozygous individuals in the population is 32/100 that equals 0.32 which is the genotipic proportion for these endividuals.

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example, the r-6 allelic frequency is 0,2. This means that if r-6=0.2, then the other allele frequency (R) is=0.8, and the sum of both the allelic frequencies equals one. This is:

p + q = 1

r-6 + R = 1

0.2 + 0.8 = 1

Then, the genotypic proportion for the homozygous individuals RR is 0.8 ² = 0.64

The genotypic proportion for the homozygous individuals r-6r-6 is 0.2² = 0.04

And the genotypic proportion for heterozygous individuals Rr-6 is 2xRxr-6 = 2 x 0.8 x 0.2 = 0.32

5 0

3 years ago

Why are skin cells undergoing mitosis continuously

marta [7]

Our skin cells divide rapidly in order to maintain a protective barrier against infection.The outer skin layer is called the epidermis and contains mostly dead cells that contain keratin.Keratin is a tough fibrous protein found in hair,skin and calluses and acts as a protective coating.The epidermis cells are constantly undergoing mitosis so the outer dead cells containing keratin are rapidly replaced as they fall off, which occurs after so many days.Then the newer living cells start producing keratin,lose their cellular contents,then die and the cycle goes on.The skin acts as a barrier to infection,therefore,it's constantly renewed and repaired.

3 0

3 years ago

The diagram shows a simple lipid.

KatRina [158]

Phospholipid is the answer

4 0

2 years ago

Read 2 more answers

Name 5 types of relationship in ecosystems.

BigorU [14]

Answer:

competition, predation, commensalism, mutualism and parasitism.

Explanation:

6 0

2 years ago

Read 2 more answers

In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fa

Dimas [21]

Answer:

See the answers below

Explanation:

Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):

(a) 9/16 D_E_ wild type

3/16 D_ee wild wing, ebony body

3/16 ddE_ dumpy wing, wild body

1/16 ddee dumpy wing, ebony body

(b) Chi square $X^2$ = $\frac{(O - E)^2}{E}$ , where O = observed frequency and E = expected frequency.

Phenotype O E $X^2$

Wild type 473 9/16 x 830 = 466.875 $\frac{(473 - 466.875)^2}{466.875}$ = 0.08

wild w/ebony b 156 3/16 x 830 = 155.625 $\frac{(156 - 155.625)^2}{155.625}$ = 0.0009

Dumpy w/wild b 149 3/16 x 830 = 155.625 $\frac{(149 - 155.625)^2}{155.625}$ = 0.28

dumpy w/ebony b 52 1/16 x 830 = 51.875 $\frac{(52 - 51.875)^2}{51.875}$ = 0.0003

Total $X^2$ = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of $X^2$ (0.05)= 7.815

(c) Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.

6 0

3 years ago