Mendel found that yellow pea pod color (G) was dominant to green pea pod color (g), and that round seeds (W) were dominant to wrinkled seeds (w). What percentage of each of the phenotypes below are expected from the following cross? GGWw x ggww

A.
0%

B.
25%

C.
33%

D.
50%

E.
66%

F.
75%

G.
100%

Yellow and wrinkled

Green and round

Green and wrinkled

Yellow and round

2. You ran an experiment in lab using four different restriction enzymes, each with a different sized recognition sequence. You used a 4-cutter, 6-cutter, 8-cutter, and 10-cutter. Your resulting gel looked like the one above. Which well contains the 8-cutter?
1. Mendel found that yellow pea pod color (G) was

A.
2

B.
3

C.
4

D.
5

1)    50% of the progeny will be yellow and rounded, G-W-

      50% of the progeny will be yellow and wrinkled, G- ww

      0% of the progeny will be green and rounded, ggW-

      0% of the progeny will be green and wrinkled, ggww

2) The well 5 will contain the 8-cutter

Explanation:

1) Available data:

Cross: GGWw x ggww

Parental)  GGWw     x     ggww

Gametes) GW GW Gw  Gw  

                 gw   gw   gw   gw

Punnet Square)    GW        GW       Gw        Gw

                 gw    GgWw    GgWw   Ggww   Ggww

                 gw    GgWw    GgWw   Ggww   Ggww

                 gw    GgWw    GgWw   Ggww   Ggww

                 gw    GgWw    GgWw   Ggww   Ggww

F1) 8/16 = 4/8 = 2/4 = 1/2 GgWw

     8/16 = 4/8 = 2/4 = 1/2 Ggww

    50% of the progeny will be yellow and rounded, G-W-

     50% of the progeny will be yellow and wrinkled, G- ww

      0% of the progeny will be green and rounded, ggW-

      0% of the progeny will be green and wrinkled, ggww

2) The well 5 will contain the 8-cutter

1) The percentages of each of the phenotypes are ;

2) The well that contains the 8 cutter is ; ( D ) Well number 5

Given Data:

Yellow pea pod ( G ) Dominant trait

green pea pod ( g ) recessive trait

Round seeds ( W ) dominant trait

wrinkled seeds ( w ) recessive trait

1) Determining the percentage of each Phenotype

Parent phenotypes to be crossed are :

      GGWw * ggww

Gametes from parents to be crossed using Punnett square

          ( GW    GW   Gw   Gw ) *  ( gw   gw   gw   gw )  

 Result from crossing :

   GgWw    GgWw   Ggww   Ggww

   GgWw    GgWw   Ggww   Ggww

   GgWw    GgWw   Ggww   Ggww

   GgWw    GgWw   Ggww   Ggww

From the Punnett square result

Number of GgWw  = 8

Number of Ggww  = 8

Total number = 16

From the Punnett square result above

percentage of GgWw = 8 / 16 * 100 = 50%  

percentage of Ggww  = 8/16 * 100 = 50%

percentage of  ggww  = 0/16 * 100 = 0%

percentage of Ggww = 0/16 *100 = 0%

Hence we can conclude that the percentage of GgWw = 8 / 16 * 100 = 50%  percentage of Ggww  = 8/16 * 100 = 50%, percentage of  ggww  = 0/16 * 100 = 0%, percentage of Ggww = 0/16 *100 = 0% and the well that contains the 8 cutter is  Well number 5.

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yes you are doing it right great gob

explanation:

answer: glucoamylase presented optimum activity at temperature of 55â°c, and, in the glycosidic bonds from the non-reducing ends of starch, resulting in the production of glucose. source and type of fermentation can affect the enzyme's properties

credit to kalahira