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3 years ago

Can i have some help plz

Biology

2 answers:

Veseljchak [2.6K]3 years ago

3 0

Answer:

ok tell me

Explanation:

i can edit this so once you tell me i can change it

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

with what

Explanation:

is it math or English

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You are given two true-breeding groups of gerbils. The gerbils in the first group have black fur and curly tails; the gerbils in

Alexxx [7]

Answer:

3 black gebrils have straight tails

Explanation:

Given -

First group of true breeding gebrils have black fur and curly tails

Second group of true breeding gebrils have yellow fur and straight tails

After the F1 generation, all the offspring would have black fur and curly tail

This means black fur is dominant over yellow fur and curly tail is dominant over straight tails

Let "Black fur" be represented by "B" and "yellow fur" be represented by "b"

Let curly tail be represented by "C" and straight tail be represented by "c"

Then parents genotype for F1 generation would be "BBCC" and "bbcc"

and genotype of first generation offspring would be "BbCc"

BbCc

BC Bc bC bc

BC BBCC BBCc BbCC BbCc

Bc BBCc BBcc BbCc Bbcc

bC BbCC BbCc bbCC bbCc

bc BbCc Bbcc bbCc bbcc

Black Curly – BBCC, BBCc, BbCC, BbCc, BBCc, BbCc, BbCC, BbCc, Bbcc,

Yellow curly- bbCC, bbCc, bbCc

Black straight – BBcc, Bbcc, Bbcc

Yellow straight- bbcc

3 black gebrils have straight tails

6 0

3 years ago

Answer please thank you!

xxTIMURxx [149]

Answer:

7. 4

8. 2

9. 3

Explanation:

Just a guess, please double check :)

Hope this helped

4 0

3 years ago

(c) Based on the data in Table 1, describe why the dominant alleles for body color and wing shape are the alleles that produce a

Brrunno [24]

Answer:

Check the explanation

Explanation:

Consider the below table of F2 data of phenotypic flies: we can reconstruct the table as seen in the second attached image below:

From the above observations it is clear the Gray body long wings and Ebony body vestigial wings are the parental combinations and that the Gray body vestigial wings and Ebony body long wings are the recombinants.

Also the ratio indicates that Gray body is dominant over ebony body and that long wings is dominant over vestigial wings.

here the genes for the two characters have shown independent assortment which means that the genes are unlinked if located on the same chromosome or are located on different chromosomes.

Now F1 hybrid= GgLl (G for Grey and L for Long)

Cross between F1 hybrid and true breeding Gray vestigial (GGll)

GgLl x GG ll

Gametes-----------> GL Gl gL gl Gl

GL Gl gL gl

Gl GGLl GGll GgLl Ggll

(Gray long) (Gray vestigial) (gray Long) (Gray vestigial)

Therefore the probability of getting the flies with gray body vestigial wings= 2/4= 50%

b) The reason why the students obtained the above given F2 results involving a cross between true breeding Gray body and long wings with true breeding Ebony body and vestigial wings is that the genes for the two charcters asssort independently in F2 generation and that the genes are not linked as:

Parents------------------> GGLL x ggll

Gametes -----------------> GL gl

F1---------------------> GgLl (Gray long but in heterozous condition)

Now GgLl x GgLl

Gametes GL Gl gL gl GL Gl gL gl

Here gametes assorted independently and hence in F2 generation we got the above results (U can show the results in the form of punnett square.