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

Answers

Answer 1

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 = 0.1056

Explanation:

Available data:

R-S are 35 mu apartCross: 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) = 0.325rs  ⇒ Parental ⇒ 32.5% (65% / 2) = 0.325Rs  ⇒ Recombinant  ⇒ 17.5% (35% / 2) = 0.175rS  ⇒  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 = 0.1137RS/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 = 0.0306

F1 Phenotypes and Proportions

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

Answer 2

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 = 0.175

In 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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AExplanation:

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Answers

Answer:

B) It is heterotrophic and unicellular.

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Answer:

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Answers

Answer:

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Answer:

Explanation:

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Explanation:

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Answer:

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Explanation:

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Answer:

Three Protons

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Answer:

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Explanation:

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Answer:

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Explanation:

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Answers

Answer:

    2/16  rounded not smelly females, RRX+X-

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Explanation:

Available data:

RR round shellrr square shellRr Pointed shellX+ is dominant and codes for not smellyX- is recessive and codes for Smelly

Cross: A pointed smelly shelled male with a round  heterozygous non-smelly female

Parentals) Rr X-Y    x     RRX+X-

Gametes) RX-, RY, rX-, rY

                RX+, RX+, RX-, RX-

Punnett square)    RX-            rX-           RY           rY

                 RX+    RRX+X-    RrX+X-    RRX+Y    RrX+Y

                 RX+    RRX+X-    RrX+X-    RRX+Y    RrX+Y

                 RX-     RRX-X-     RrX-X-     RRX-Y     RrX-Y

                 RX-     RRX-X-     RrX-X-     RRX-Y     RrX-Y

F1) 2/16 bivalves are expected to be rounded not smelly females, RRX+X-

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    2/16 bivalves are expected to be rounded smelly females, RRX-X-

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    2/16 bivalves are expected to be rounded not smelly males, RRX+Y

    2/16 bivalves are expected to be pointed not smelly males, RrX+Y

    2/16 bivalves are expected to be rounded smelly males, RRX-Y

    2/16 bivalves are expected to be pointed smelly males, RrX-Y

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Answer:

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IM BiuLt  DIFFRENT

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Answers

Answer:

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Answers

Answer and Explanation:

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Answers

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Answers

Answer:

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Explanation:

....

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Answers

Answer:

Explanation:

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Answer:

1. uterus is also known as the womb.

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Answers

Answer:

True

Explanation:

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Answers

Answer:

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Answers

Answer:

4 percent (4%)

Explanation:

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Please help1!1

Answers

Answer:

D

Explanation:

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Answers

Answer:

attacks the outermost phosphorous group of the incoming nucleotide.  

Explanation:

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Answers

d is correct

..................

^_^

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