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katen-ka-za [31]
3 years ago
6

The barley plant has seven pairs of chromosomes. For a gamete, the number of genetic combinations that are possible through inde

pendent assortment is _____. For a zygote produced by two barley parents, the number of possible genetic combinations is ________.
Answers:
1. 128
2. 16,384

I hope this helps.

Biology
1 answer:
skad [1K]3 years ago
4 0

Answer:

1. 128

2. 16,384

Explanation:

Imagine there were two pairs of chromosomes:

A paired with B

C paired with D

A gamete could have the following combinations:

A + C

A + D

B + C

B + D

Therefore, with two pairs of chromosomes, there are 4 (2²) possibilities

With seven pairs of chromosomes, there are  (2⁷) possibilities. 2⁷  = 128

For a zygote produced by two barley parents, there are two gametes fused. So we must multiply the number of combinations for each gamete. Therefore, number of possible genetic combinations is 128 x 128 = 16,384

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This is for the table at the top

<em>Plentiful roaming space</em>- increase - <u>because they</u><u> have more space they can live in.</u>

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<em>Habitat </em><em>destruction</em><em> </em><em>-</em><em> </em>decrease - <u>Because they won't have a place to live in </u>

<em>Elimination of predators</em><em> </em><em>-</em><em> </em>increase - <u>Because there will be less animals hunting them for food.</u>

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3 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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