Aardvark Eye Color The trait works following the rules overof codominance. Brown

Question

Aardvark Eye Color
The trait works following the rules overof codominance. Brown (B)is dominant over gold (b).

Aardvark Snouts
Blunt Snouts (N) are dominant over pointy snouts (n). Snout shapeis controlled by alleles that interact following the rules ofsimple dominance

If a gold eye, blunt nose (in this case heterozygous) breeds with abrown and gold eyed( in this case heterozygous), blunt nose (alsoheterozygous) what will the kids look like

I got this far i don't know how to describe the phenotypicratios.
bbNn x BbNn

BN Bn bN Bn bN BbNN BbNn bbNN bbNn bn BbNn Bbnn bbNn bbnn bN BbNN BbNn bbNN bbNn bn BbNn Bbnn bbnn bbnn

BN Bn bN Bn bN BbNN BbNn bbNN bbNn bn BbNn Bbnn bbNn bbnn bN BbNN BbNn bbNN bbNn

Explanation / Answer

I see where your confusion is, I am sorry I gave you the wronginformation When I saw your cross I automatically assumed that is was straightdominance but now that I have re-read the question I see that weare dealing codominance. codominance - is when neither one trait nor the other are dominantand it leads to offspring that show traits of both, like in yourexample (if your offspring has both alleles then they will haveboth traits, the only way to have an offspring with one or theother trait is to be homozygous for that particular trait)The error in your square is the representation of thealleles when one trait is not dominant over the other the trait thealleles are not represented as capital and lower case (ex. B and b)they are represented each with a capital letter but one of them hasa prime mark (ex. B and B') incomplete dominance - is when you have to alleles that haveoffspring with a completely different trait than the parents (forexample a white and a red flower yielding offspring that arepink) So I started the problem from scratch just to double check andreworked it for you: Let's start by representing our alleles: N = Blunt (dominant) n = pointy (recessive) B = Brown B' = Gold The parents are gold eye/blunt nose (hetero) and a brown and goldeye/blunt nose (hetero) so we know that both parents are blunt nose hetero yielding =Nn one parent is gold eyed B'B' (must be homo because it's the onlyway to have that trait) one parent is brown and gold eyed BB' (must be heterozygous) cross of the parents traits yield                                                      B'                                   B' B' B'B' B'B' B B'B B'B'                                                       N                           n N Nn Nn n Nn nn                                 B'B'                B'B'                 B'B               B'B' NN B'B'NN B'B'NN B'BNN B'B'NN Nn B'B'Nn B'B'Nn B'BNn B'B'Nn Nn B'B'Nn B'B'Nn B'BNn B'B'Nn nn B'B'nn B'B'nn B'Bnn B'B'nn so we have B'B'NN = 3 = gold/blunt B'B'Nn = 6 = gold/blunt B'B'nn = 3 = gold/pointy B'BNn = 2 = gold/brown/blunt B'BNN = 1 = gold/brown/blunt B'Bnn = 1 = gold/brown/pointy add up the traits that are similar to get the phenotypic ratio gold/blunt = 3 + 6 = 9 gold/pointy = 3 gold/brown/blunt = 2 + 1 = 3 gold/brown/pointy = 1 so your ratio is 9:3:3:1 for phenotypes I hope this helps, if you are still unsure please email me!

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