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Delvig [45]
3 years ago
6

Garden plants are eaten by snails, which are eaten by birds, which are eaten by the household cat. What happens to the other mem

bers of the food chain if the cat is removed? a. The number of plants, snails, and birds increase. b. The number of plants and snails increase and the number of birds decreases c. The number of plants increases and the number of snails and birds decrease. d. The number of plants and birds increase and the number of snails decreases.
Biology
2 answers:
Alex17521 [72]3 years ago
8 0

The correct answer is (d) The number of plants and birds increases and the number of snail decreases.

Green plants are eaten by snails, snails by birds, and birds by household cat. If cat is removed from the cycle then the number of birds will increase which will decrease the number of snail in the environment.  As there are large number of birds there will be a great demand of snails as a food for birds. Then there will be a decrease in the number of snails because of over eating of snails. There will be an increase in the number of garden plants because very less number of snails are left that will eat garden plants.

Lerok [7]3 years ago
4 0

The correct answer is d. The number of plants and birds increase and the number of snails decreases.

The plants are eaten by the snails, the snails by the birds and birds by the cats. If the cats are removed from this food chain then there will be more number of birds surviving hence the number of birds increase. The birds will eat more snails so the number of snails will decrease. When the number of snails decreases, they will less number of plants therefore the number of plants increases. Every tropic level in the food chain has its own different impact on the number of species of the food chain.

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Answer:

I think they work together to create tissue

Explanation:

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Which is true about the inside of an animal cell?
andrey2020 [161]

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Where are the choices ?

Explanation:

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3 years ago
In the family tree below, people with the recessive trait of attached earlobes are shaded gray
serious [3.7K]
<h2><u>Full Question:</u></h2>

In the family tree below, people with the recessive trait of attached earlobes are shaded gray.

What must be true about the person labeled "A"?

A. It is a male with at least one dominant allele.

B. It is a male with two dominant alleles.

C. It is a female with at least one dominant allele.

D. It is a female with two dominant alleles.

<h2><u>Answer:</u></h2>

Its a male with atleast one dominant allele.

Option A.

<h3><u>Explanation:</u></h3>

The gene for the attached earlobe is recessive while the gene for the free earlobes is dominant. In the phylogenetic tree, we can see that both the father and mother aren't having attached earlobes. So both of them are having atleast one dominant allele which makes them have free earlobe.

In the F1 offsprings, one of the female and a male is having free earlobes. So both of them have atleast one dominant allele. The 2nd female is having an attached earlobe. So both the recessive allele have come form one parent each. So both of them are heterozygous.

Thus, the male marked as A atleast have one dominant allele. He can be a homozygous dominant, but the probability is 25%.

8 0
3 years ago
4. The Rh blood type response system is controlled by the D allele. The genotypes DD and Dd are Rh (Rh positive); dd is Rh- (Rh
DedPeter [7]

Answer:

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.

EXPLANATION:

The equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1 - This is according to Hardy-Weinberg.

So p+q =1

p2+ 2pq+q2 =1

where p = frequency of dominant allele

q is the frequency of the recessive allele.

from the question, it was given that there were  170 individual out of 400, which were Rh- negative,

So q2 = 170/400 = 0.425

q= 0.65

Also p+q =1

so p = 1-q

or p = 1-0.65

Hence p =0.35

Frequency of homozgupus for D allele = 0.35*0.35 = 0.1225

Frequency of heterozygous or Dd will be 2pq.

or 2*0.35*0.65 = 0.455

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.

8 0
4 years ago
Read 2 more answers
Explain what terms in the Hardy-Weinberg equation give: allele frequencies (dominant allele, recessive allele, etc.) each genoty
zysi [14]

Answer:

Explanation:

Hardy Weinberg equation is presented below and describes that in a population the frequency of alleles ad genotypes will remain static or the same in the absence of evolutionary disturbances such as mutation, migration ( gene flow), natural selection and with the population large and random mating

p² +2pq + q²

where p represents the frequency of the dominant alleles

q represent the frequency of the recessive alleles

p² represent the frequency of the dominant homozygous genotype

q² represent the frequency of the recessive homozygous genotype

2pq represent the frequency of the heterozygous genotype

q² also represent the frequency of the recessive phenotype

(p² + 2pq) represent the frequency of the dominant phenotype

5 0
3 years ago
Read 2 more answers
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