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 proportions
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 = 0.1443rr/S- = 0.1443rr/ss = 0.1056Explanation:
Available data:
R-S are 35 mu apartCross: RS/rs x RS/rsFirst, 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) = 0.325rs ⇒ Parental ⇒ 32.5% (65% / 2) = 0.325Rs ⇒ Recombinant ⇒ 17.5% (35% / 2) = 0.175rS ⇒ Recombinant ⇒ 17.5% (35% / 2) = 0.175F1 Genotypes and Proportions
RS/RS = 0.325 x 0.325 = 0.1056 RS/Rs = 2x 0.325 x 0.175 = 0.1137RS/rS = 2x 0.325 x 0.175 =0.1137 rs/rs = 0.325 x 0.325 = 0.10562rs/RS = 2x 0.325 x 0.325 = 0.2112rs/Rs = 2x 0.325 x 0.175 = 0.1137rs/rS = 2x 0.325 x 0.175 = 0.1137Rs/rS = 2x 0.175 x 0.175 = 0.061Rs/Rs = 0.175 x 0.175 = 0.0306rS/rS = 0.175 x 0.175 = 0.0306F1 Phenotypes and Proportions
R-/S- = 0.1056 + 0.1137 + 0.1137 + 0.2112 + 0.061 = 0.6052 R-/ss = 0.1137 + 0.0306 = 0.1443rr/S- = 0.1137 + 0.0306 = 0.1443rr/ss = 0.1056
If R and S loci are 35 m.u. apart and a plant of genotype RS/rs is selfed, then the progeny phenotypes will be 0.606 RS; 0.106 rs; 0.144 Rs and 0.144 rS.
In this case, it is imperative to estimate gamete frequencies
RS = 0.65% / 2 = 0.325rs = 0.65% / 2 = 0.325Rs (recombinant) = 0.35 / 2 = 0.175rS (recombinant) = 0.35 / 2 = 0.175In consequence, the F1 genotypes will be RS/RS = 0.1056 (i.e., 0.325 x 0.325); RS/Rs = 0.1137; RS/rS = 0.1137; rs/rs = 0.10562; rs/RS = 0.2112; rs/Rs = 0.1137; rs/rS = 0.1137; Rs/rS = 0.06; Rs/Rs = 0.0306; and rS/rS = 0.0306.Finally, it is possible to obtain phenotypic frequencies by summing genotypic frequencies for each case in particular.In conclusion, if R and S loci are 35 m.u. apart and a plant of genotype RS/rs is selfed, then the progeny phenotypes will be 0.606 RS; 0.106 rs; 0.144 Rs and 0.144 rS.
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Answer:
Radio waves
Explanation:
kase mas mababa Ang secondary kesa sa tertiary
Please answer the questions
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A.
During an earthquake, both P waves and S waves travel away from the focus through the interior of Earth.
B.
After an earthquake, P waves are detected on the opposite side of Earth but S waves are not.
C.
P waves do not refract based on the density of the medium it is traveling in, but S waves do.
D.
Seismographs that are far from the epicenter of an earthquake detect P waves much sooner than S waves.
Explanation:
During
an earth quake both p waves and s waves travel away from the focus
Answer:
B
Explanation:
B
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Answer:
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Explanation:
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Answer:
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Answer:
4 percent (4%)
Explanation:
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