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Daniel [21]
3 years ago
6

In humans, long eyelashes (L) are dominant to short eyelashes (l) and wet ear wax (W) is dominant to dry ear wax (w). Assume the

genes for eyelash length and ear wax consistency are on different chromosomes and assort independently. Two heterozygous individuals meet on match and have a child. What is the chance that the child has long eyelashes and dry ear wax?
Biology
2 answers:
Alina [70]3 years ago
6 0

Answer:

<h2>3/16</h2>

Explanation:

Given:

Long eyelashes (L) are dominant to short eyelashes(l), So L is dominant on l.

Wet ear wax (W) is dominant  dry ear wax (w) , so W is dominant on w.

Two heterozygous parents are...

LlWw and LlWw

F1 generation

3 child have long eyelashes and dry wax = L_ww .

There is 3/16 ( 3 out of 16) chances that the child have long eyelashes and dry ear wax and their genotype is  L_ww.

disa [49]3 years ago
6 0

Answer:

2 different types of genes are given,

long lash (L) and short lash (l)

wet ear wax (W) and dry ear wax ( w)

here two heterozygous individuals are crossed

LLww × llWW

         ↓

     LlWw    ( F1 gen)

         ↓

 (1)  LL  (2) Ll   (1) ll    [ as 1:2:1 ( 1= dominant and recessive but 2= hybrid)]

(1) WW (2)  Ww (1) ww

          ↓

1 LLWW   2 LLWw   1 LLww

2 LlWW  4 LlWw   2 Llww

1 llWW     2 llWw    1 llww

therefore,

long lash and dry ear wax = LLww (1) + Llww (2)

                                      =  3/16

there is a 3/16 chance that the child will have long eyelash and dry ear wax.

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