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Lilit [14]
3 years ago
5

How do introduced species become established in a new ecosystem? Check all that apply. introduced to a habitat similar to their

own introduced to a habitat different than their own outcompete native species generally have no native predators generally have native predators often have high reproductive rates often have low reproductive rates can tolerate a range of conditions
Biology
2 answers:
raketka [301]3 years ago
4 0
. out compete native species. can tolerate range conditions. high reproduction rates
NISA [10]3 years ago
4 0

Answer:

1, 3, 4, 6, 8

Explanation:

just took the test. :D

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Which of the following could be a tertiary consumer?
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Which group is allowed to unionize but NOT strike? *
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Please help with this
Tanya [424]
Test One: All short fur
Phenotype ratio: 4:0
Test Two: three short fur and one long fur
Phenotype ratio: 3:1
Test Three: all long fur
Phenotype ratio: 4:0
Genotypes for fur length dominant
G: FF, and Ff (These are the genotypes for short fur, short fur is a dominant gene.
P: Short fur is dominant
Genotypes and phenotypes for fur length recessive
G: ff
P: Long fur is recessive
If you have a hamster with short fur, the possible genotypes the hamster could have is FF, and Ff
If you have a hamster with long fur, the possible genotype is ff.
5 0
3 years ago
Suppose the human trait for hair type is controlled by a simple dominant and recessive relationship at one locus. Curly hair, C,
myrzilka [38]

Answer:

A) The frequency of the dominant allele, C = 0.4376

B) frequency of the heterozygous genotype, Cc = 0.2461

Explanation:

Given -

Curly hair, C, is the dominant allele, and straight hair, c, is the recessive allele

Number of students with curly hair is 37

Thus frequency of genotype "CC"

= q^{2} = \frac{37}{117} \\= 0.3162

Thus, frequency of allele "q"

= \sqrt{0.3162} \\= 0.5623

As per Hardy Weinberg's first  equation-

p+q=1

Substituting the value of q, we get -

p+0.5623=1\\p= 0.4376

As per Hardy Weinberg's second equation -

p^2+q^2+2pq=1\\

Substituting the values of "p" and "q" we get -

0.4376^{2} + 0.3162+2pq = 1\\2pq = 0.2461\\

Hence,

A) The frequency of the dominant allele, C = 0.4376

B) frequency of the heterozygous genotype, Cc = 0.2461

8 0
3 years ago
Read 2 more answers
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