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levacccp [35]
3 years ago
7

How are fished adapted to their environment?

Biology
1 answer:
Rainbow [258]3 years ago
7 0
They find cirten spots and food and fishes to adapt with 
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Using the product rule of probability, calculate the odds of getting an individual that receives the bb genotype and therefore s
Pepsi [2]

Answer:

B) 1/2 X 1/2 = 1/4

Explanation:

It happens when both the parents are heterozygous for the trait. The cross related to this result is shown as under:

                          Father           Mother

                             Bb       x        Bb

                             /   \                 /   \

Gametes:           B      b            B      b

                          |        |              |       |

Probability:      1/2     1/2          1/2    1/2

The probability of formation of 'b' gamete from father is 1/2 because there are only two gametes 'B' and 'b' and out of these two one will be assorted as 'B' and another one as 'b' and from mother also the probability of formation of 'b' gamete is 1/2.

Now the assortment of gametes with each other is an independent event i.e. any gamete from father can fuse with any gamete of mother so the overall probability of formation of 'bb' genotype will be 1/2 x 1/2 = 1/4.

3 0
3 years ago
This is the process in which two chromosomes exchange DNA during prophase of meiosis.
Sergio039 [100]
Crossing over between two chromosomes
8 0
3 years ago
Read 2 more answers
In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The gene
NikAS [45]

Answer:

A) 9:3:3:1

B) All bitter fruit, yellow spotted offsprings

C) Phenotypes are bitter yellow spotted (4), bitter no spot (4), sweet yellow spot (4), and sweet no spot (4). 1:1:1:1

Explanation:

This is a typical dihybrid cross involving two genes, one coding for fruit taste and the other for spot color. The allele for bitter taste (B) and yellow spot (S) is dominant over the allele for sweet taste (b) and no spot (s) respectively.

Hence, a heterozygous F1 resulting from a cross between an homozygous dominant (bitter fruit, yellow spot) and homozygous recessive (sweet fruit, no spot) will have a BbSs genotype. The heterozygous F1 offsprings are self-crossed and produce gametes BS, Bs, bS, bs. (See punnet square). The F2 offsprings will have the following phenotypes: Bitter fruit, yellow spot (9)

Bitter fruit, no spot (3)

Sweet fruit, yellow spot (3)

Sweet fruit, no spot (1)

Back cross between a F1 offspring (BbSs) and homozygous dominant parent (BBSS) will produce all bitter fruit, yellow spot offsprings (see attached image). BBSS (4), BBSs (4), BbSS (4), and BbSs (4) are the offsprings' genotypes.

For the back cross between a F1 offspring (BbSs) and a homozygous recessive (bbss) parent, the Phenotypes with their proportions are as follows:

Bitter fruit, yellow spot (BbSs, 4)

Bitter fruit, no spot (Bbss, 4)

Sweet fruit, yellow spot (bbSs, 4)

Sweet fruit, no spot (bbss, 4).

4 0
3 years ago
What are the products of semiconservative replication for a double-stranded dna molecule?
Triss [41]
The products of semiconservative replication for a double stranded DNA molecule are two double stranded DNA molecules, one side of both strands being the old strands that were separated and used as a template and the other side of both strands being the nucleotides that were placed on the template strand.
7 0
4 years ago
If the frequency of the black allele %B allele for the moth coloration gene in the newly founded island population of moths in y
deff fn [24]

Answer:A

Explanation:

7 0
3 years ago
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