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skelet666 [1.2K]
3 years ago
10

Step 5: use the data to model the sizes of the flocks in the second generation

Biology
1 answer:
Illusion [34]3 years ago
5 0

Answer:

Flock X Food Percentage: 41%,

Simulated Number of Birds in Flock for 2nd Generation**: 12

Flock Y Food Percentage: 33℅

Simulated Number of Birds in Flock for 2nd Generation**: 10

Flock X Food Percentage: 26%

Simulated Number of Birds in Flock for 2nd Generation**: 8

Explanation: * Divide each flock's total pieces of food by 300, the total number of pieces of food eaten.

** Multiply the food percentage for each flock by the total number of birds (30).

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What does meiosis and mitosis have in common
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The same general processes occur in meiosis and mitosis 
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how many people most likely represents a snapshot in time of a human society that existed 12,000 years ago?
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The surgical placement of an artificial joint is known as
Alekssandra [29.7K]

Answer:

Arthroplasty

Explanation:

The restoration of function of a joint is called Arthroplasty. The bone functions are restored by maintaining the  bones through resurfacing. Some times artificial joints are also used instead of resurfacing the existing bones.

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6 0
3 years ago
In a plant, leaf color and leaf shape are controlled by two linked genes Leaves of the wild-type plant are red. A recessive muta
Brrunno [24]

Solution :

Red is dominant over the white leaves and pointed is dominant over smooth.

Red = R ,    white = r

Pointed = P ,   smooth = p

Red, pointed x white,smooth ------ Parents

RRPP                 rrpp

RP                      rp  ----------------- Gametes

         RrPp    -------- $F_1$

  (red, pointed)

When this $F_1$ is test crossed,

RrPp  x  rrpp     ----  $F_1$ test cross

Gametes   → rp

RP ---------- RrPp - red, pointed - 36 parental type

Rp ---------- Rprp - red, smooth - 14 re- combination

rP ----------- rrPp - white pointed - 10 re- combination

rp ----------- rrpp - white, smooth - 40 - parental type

                                    Total =   100

As this ratio is deviating from 1:1:1:1, it indicated the two genes are linked.

Linkage strength = percentage of crossing over = map distance between the genes.

Percentage crossing = recombination frequency = percentage of recombination.

Recombination frequency $=\frac{\text{total no. of recombination}}{\text{total no. of progeny}}\times 100$

                                         $=\frac{14+10}{100} \times 100$

                                         = 24 %

                                        = 24 map units

4 0
3 years ago
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