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Variation in the appearance of guppy color patterns to guppies and their predators under different visual conditions.

kompoz [17]

Color patterns can be thought of as mosaics of different-sized, form, brilliant, and colored patches. A number of natural conditions, including ambient light, watercolor, viewing distance, and the visual capabilities of guppies and three of their predators, were used to assess how noticeable guppy color patterns were.

A particular color pattern element (patch reflectance )'s spectrum, the medium's transmission spectrum, and the ambient light spectrum that strikes the patch all interact to determine the light spectrum that reaches the eye's surface (air or water).

The light at distance x from the patch that reaches the eye at any given wavelength, I, is the result: Q(& x) is the photon flux from the patch reaching the eye (patch radiance), and Q(A) is the photon flux of ambient light incident on the patch.

Predation intensity varies spatially with guppy color patterns, which are influenced by the make-up of the predator communities (Endler, 1978). As a result, calculations were made independently for the color patterns of guppies that are unique to three different predator communities.

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2 years ago

Which part of the earth system contains continents?

Olenka [21]

Continents are land masses that exist on the surface of the earth. The continents are part of the uppermost layer of the earth which is called the crust.

6 0

3 years ago

A scientist hypothesizes that the temperature at which an alligator's egg is incubated will determine whether the alligator will

lys-0071 [83]

Answer:

9

Explanation:

8 0

3 years ago

True or False?

Montano1993 [528]

Answer:

False

Explanation:

Goold luck and have a great day c:

6 0

3 years ago

The R and S loci are 35 m.u. apart. If a plant of genotype is selfed, what progeny phenotypes will be seen and in what proportio

Veronika [31]

Complete question:

The R and S loci are 35 m.u. apart. If a plant of genotype RS/rs is selfed, what progeny phenotypes will be seen and in what proportions

Answer:

R-/S- = 0.6052
R-/ss = 0.1443
rr/S- = 0.1443
rr/ss = 0.1056

Explanation:

<u>Available data:</u>

R-S are 35 mu apart
Cross: RS/rs x RS/rs

First, we need to recognize the parental gametes and the recombinant ones.

Cross:

Parentals) RS/rs x RS/rs

Gametes) RS Parental ⇒ Equal to the parental genotype

rs Parental ⇒ Equal to the parental genotype

Rs Recombinant ⇒ Product of recombination

rS Recombinant ⇒ Product of recombination

We know that genes are 35 MU apart from each other. The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

According to this information, if 1 MU = 1% recombination frequency, then

35MU --------- 35% recombination frequency = 0.35

Now, let us calculate the frequency of each gamete type.

35 map units = 35 % of recombination in total

= % Rs + % rS

= 17.5% Rs + 17.5% rS

Now, if the recombination frequency equals 35% then the parental frequency is 100% - 35% = 65%

65 % parental frequency = % of RS + % rs

= 32.5% RS + 32.5% rs

The frequency of each gamete is

RS ⇒ Parental ⇒ 32.5% (65% / 2) = 0.325
rs ⇒ Parental ⇒ 32.5% (65% / 2) = 0.325
Rs ⇒ Recombinant ⇒ 17.5% (35% / 2) = 0.175
rS ⇒ Recombinant ⇒ 17.5% (35% / 2) = 0.175

F1 Genotypes and Proportions

RS/RS = 0.325 x 0.325 = 0.1056
RS/Rs = 2x 0.325 x 0.175 = 0.1137
RS/rS = 2x 0.325 x 0.175 =0.1137
rs/rs = 0.325 x 0.325 = 0.10562
rs/RS = 2x 0.325 x 0.325 = 0.2112
rs/Rs = 2x 0.325 x 0.175 = 0.1137
rs/rS = 2x 0.325 x 0.175 = 0.1137
Rs/rS = 2x 0.175 x 0.175 = 0.061
Rs/Rs = 0.175 x 0.175 = 0.0306
rS/rS = 0.175 x 0.175 = 0.0306

F1 Phenotypes and Proportions

R-/S- = 0.1056 + 0.1137 + 0.1137 + 0.2112 + 0.061 = 0.6052
R-/ss = 0.1137 + 0.0306 = 0.1443
rr/S- = 0.1137 + 0.0306 = 0.1443
rr/ss = 0.1056

4 0

3 years ago