I think itts virchow
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Answer:
<h2>(14+16)/1000 *100= 3</h2>
Explanation:
Parental combinations are always more common then other recombinant progenies.
A female fly is heterozygous for three X-linked genes crossed with an abc/Y male. . If single crossover occurs between a & b and between b & c, then the double cross is that, which occurs in both.
Double cross over progeny are always least in number, so here double cross occurs are in abC 14 and ABc 16.
So we are expecting that number of double cross over progeny= (14+16)/1000 *100= 3
The probability of a crossover occurring between two particular genes on the same chromosome increases as the distance between those
If you are implying one these choices not needed for survival it would be parents only when you can survive on your own (I'm not that sure but I hope this helped the slightest but)
Answer:
Yellow smooth - 9
Yellow wrinkle - 3
Green smooth- 3
Green Wrinkle - 1
Explanation:
Let the green color of the seed be depicted by "G" and the yellow color of the seed be depicted by "g"
Let the smooth the seed be depicted by "R" and wrinkled seed be depicted by "r"
F1 cross -
true breeding smooth green plant ( RRGG) and true breeding wrinkled yellow (rrgg)
F1 gamete will be RG, RG, rg, rg
F1 offspring will be RrGg , Thus all F1 offspring will be heterozygous smooth and yellow.
Thus, R is dominant over r and g is dominant over G
F2 Generation –
RrGg x RrGg
Gametes will be RG, Rg, rG, rg
RG Rg rG rg
RG RRGG RRGg RrGG RrGg
Rg RRGg RRgg RrGg Rrgg
rG RrGG RrGg rrGG rrGg
rg RrGg Rrgg rrGg rrgg
R is dominant over r and g is dominant over G
Genotypes are –
RRGG - 1 (Smooth Green)
RRGg-2 (Smooth yellow)
RrGG-2 (Smooth Green)
RrGg-4 (Smooth yellow)
RRgg- 1 (smooth yellow)
Rrgg – 2 (Smooth yellow)
rrGG – 1 (wrinkled Green)
rrGg – 2 (Wrinkled yellow)
rrgg – 1 (wrinkled yellow)
Yellow smooth - 9
Yellow wrinkle - 3
Green smooth- 3
Green Wrinkle - 1
