need an explination for these two problem the answer for #11 is B and the answer
ID: 281005 • Letter: N
Question
need an explination for these two problem
the answer for #11 is B and the answer for the #12 is D
11.Mice with black coat color are homozygous aa for a recessive gene. Mice with agouti coat color can be AA or Aa. Mice the short tails are homozygous I. Mice with long tails can be LL or LI. If heterozygous mice that are AaLI are crossed to each other (AaLI X AaL), the expected phenotypic ratio will be (A) 9 black short: 3 agouti short: 3 black long: 1 agouti long (B) 9 agouti long: 3 agouti short: 3 black long: 1 black short (C) 1 agouti long: 1 agouti short: 1 black long: 1 black short (D) 3 agouti long: 1 black short AA or Aaagouti, aa-black LL or Ll = long. 12. In a particular plant, two genes control leaf shape and color. Round leaves (A) are dominant to jagged leaves (r). Yellow fruits (Y) are dominant to white fruits(y) . A true breeding round-leaved, yellow-fruited plant is mated with a jagged-leaved, white-fruited plant. What are the genotypes of the plants involved in this cross? (A) RRYY X RRYY (B) RrYy X rryy (C) RrYy X RrYy (D) RRYY X rryy (E) Cannot determine from RR and Rr round r jagged YY and Yy= yellow the information givernExplanation / Answer
11) The cross between mice that are heterozygous (AaLl) are shown below,
The offsprings in F2 generation are seen in the following ratio,
9:3:3:1 of agouti long, agouti short, black long, black short. So, the answer for this question is option B.
12) In plants, round leaves (R) are dominant over jagged leaves (r). Yellow (Y) fruits are dominant to white (y) fruits. True breeding round leaved (RR) and yellow fruited (YY) plant is mated with jagged leaves (rr) and white fruit (yy). So, the genotypes of the plants involved in the cross is RRYY and rryy. Therefore, the correct answer for 5his question is option D.
AL Al aL al AL AALL AALl AaLL AaLl Al AALl AAll AaLl Aall aL AaLL AaLl aaLL aaLl al AaLl Aall aaLl aallRelated Questions
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