1. Pigment in the mouse is determined to some extent by two independently assort
ID: 275769 • Letter: 1
Question
1. Pigment in the mouse is determined to some extent by two independently assorting but interacting gene loci. In the presence of the C allele pigment is produced from a colorless precursor. The allele c inhibits pigment formation entirely and gives the albino phenotype when homozygous. If the allele A of the other locus is present, pigment will be deposited and results in the agouti pattern. In the absence of A (i.e., when aa is the genotype) only black pigment pattern will occur.
What F1 and F2 genotypic ratios will be obtained from a cross between a AACC x aacc? Show all your work
P genotypes Genotypes
AACC x aacc ___________ x __________
F1 Genotype and phenotype
F2 Genotypes and Phenotypes
In the following crosses shown below between agouti females whose genotypes are unknown and males of the aacc genotype, what are the genotypes of the female parents for each of the following progeny?
8 agouti and 8 colorless
9 agouti and 10 black
4 agouti, 5 black, 10 colorless
Explanation / Answer
According to the given data, the pigment colour corresponds to these genotypes.
cc = colourless
aa = black
Aa / AA = Agouti
F1 generation after crossing AACC x aacc would result in
The resultant phenotype : All progeny of F1 generation would be Agouti coloured. And the genotype would be AaCc
F2 generation: AaCc x AaCc when crossed would result in
Genotypes and the Phenotypes obtained are:
1 AACC : 2 AACc : 2 AaCC : 4 AaCc : 1 AAcc : 2 Aacc : 1 aaCC : 2 aaCc : 1 aacc
9 Agouti (AACC, AACc, AaCC, AaCc) : 3 Black (aaCc, aaCc) : 4 Colourless (AAcc, Aacc, aacc)
Next question:
Female Agouti genotype = AACc
AACc x aacc when crossed results in 8 Agouti and 8 colourless
Female genotype = AaCC would result in equal number of Agouti and Black coloured progeny. (Kindly check the question as to how one obtained 19 progeny in an F2 cross)
Female genotype = AaCc would result in Agouti and Black coloured progeny, with Colourless progeny being double in number. (Again here, kindly check the question as to how one obtained 19 progeny in an F2 cross)
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