7. What are the all of the possible genotypes for the following human blood

Red foxes have three different types of coat color: red, cross, and black/silver. These coat colors are controlled by a single l

Anna [14]

Answer:

74%

Explanation:

The frequency in such cases (diallelic cases) is estimated by following formula which is attached in the image:

So as per our question, the frequency of the "R allele" as per data given below will be as under:

58 red-phase (RR)

32 cross-phase (RS)

10 black/silver-phase (SS) foxes

Total plants = 58 + 32 + 10 = 100

f(R) = 58 x 2 + 32/ 2 (58+32+10)

= 116 + 32 / 200

= 148/200

= 0.74 i.e. 74%

3 0

3 years ago

all but one describes a function of apoptosis. That is a) cell death that occurs when cells are exposed to toxins or are injured

natima [27]

The correct answer is A. Apoptosis is regulated cell death/suicide, not accidental.

4 0

2 years ago

Read 2 more answers

When the comb jelly, Mnemiopsis leidyi, was introduced into the Black Sea, its population exploded to 500 comb jellies per cubic

nikitadnepr [17]

The answer is invasive species.

Invasive species are species that are not native to specific ecosystem, but they have a capability to spread widely through the ecosystem, This way they can threaten native species of the ecosystem or cause environmental harm.

After being introduced into the Black Sea, they greatly affect local ecosystem. In the new area, they have no natural predators and are able to spread. If they spread beyond the area they are introduced, they become invasive species which can overgrow native biotopes and are a great threat to the local biodiversity. This could lead to the extinction of some native species.

7 0

3 years ago

In contrast to eukaryotic cells, prokaryotic cells

LenKa [72]

The answer to this is C.

7 0

3 years ago

Read 2 more answers

The Offspring produced by a cross between two given types of plants can be any of the three genotypes denoted by A, B, and C. A

ratelena [41]

Complete question:

The offspring produced between two given types of plants can be any of the three genotypes, denoted by A, B and C. A theoretical model of gene inheritance suggests that the offspring of types A, B and C should be in a 1:2:1 ratio (meaning 25% A, 50% B, and 25% C). For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded in the table below.

Genotype Observed frequency

A → 18 individuals

B → 55 individuals

C → 27 individuals

Do these contradict the genetic model?

Use a 0.05 level of significance.

Determine the chi-square test statistic.

Answer:

Do these contradict the genetic model? No, according to the chi-square test, there is not enough evidence to reject the null hypothesis of the population being in equilibrium.

Explanation:

Available data:

Crossed genotypes: two
Genotypes among the offspring; Three → A, B, and C
Expected phenotypic ratio → 1:2:1
Total number of individuals, N = 100
A = 18 individuals
B = 55 individuals
C = 27 individuals

So, let us first state the hypothesis:

H₀= the population is equilibrium for this locus → F(A) = 25%, F(B) = 50%, F(C) = 25%
H₁ = the population is not in equilibrium

Now, let us calculate the number of expected individuals, according to their expected ratio.

4 -------------- 100% -------------100 individuals

1 --------------- 25% -------------X = 25 individuals A

2 -------------- 50% -------------X = 50 individuals B

1----------------- 25%--------------X = 25 individuals C

A B C

Observed 18 55 27
Expected 25 50 25
(Obs-Exp)²/Exp 1.96 0.5 0.16

(Obs-Exp)²/Exp

A) (18 - 25)²/25 = 49/25 = 1.96

B) (55 - 50)² / 50 = 25/50 = 0.5

C) (27 - 25)²/25 = 4/25 = 0.16

Chi square = X² = Σ(Obs-Exp)²/Exp

∑ is the sum of the terms
O are the Observed individuals: 2 in chamber B, and 18 in chamber A.
E are the Expected individuals: 10 in each chamber

X² = ∑ ((O-E)²/E) = 1.96 + 0.5 + 0.16 = 2.62

Freedom degrees = 2

Significance level, 5% = 0.05

Table value/Critical value = 5.99

X² < Critical value

2.62 < 5.99

These results suggest that there is not enough evidence to reject the null hypothesis. We can assume that the locus under study in this population is in equilibrium H-W.

6 0

3 years ago