# Genetics problem, test-crossed [closed]

Image description: text reads "An $F_1$ x $F_1$ self gives a 15:1 phenotypic ratio in the F2. What phenotypic ratio would you expect if you test-crossed the F1?"

How do I find out the $F_1$ phenotypic ratio by using the given $F_2$ phenotypic ratio? What method should I use to solve?

• Please show an attempt at solving this question or it may be closed. – L.B. Oct 26 '16 at 13:27
• Test-crossed f1 : AaBb X aabb AaBb:1/4 Aabb:1/4 aaBb:1/4 aabb:1/4 ratio is 1:1:1:1 – eric pang Oct 26 '16 at 13:34
• I have no idea what should i do next – eric pang Oct 26 '16 at 13:36
• PLEASE DO NOT POST TEXT AS IMAGES. Copy and paste the text into your question. Images are not searchable, and can not be interpreted by screen readers for those with visual impairments. Use the edit link to modify your question. See this for more information. – MattDMo Oct 26 '16 at 14:13

The 15:1 phenotypic ratio resulting from the self cross suggests duplicate dominant genes without cumulative effect.
In other words, two genes with complete dominance and dominant alleles producing the same phenotype, but independently, meaning that dominance at one locus ~ dominance at the other locus ~ dominance at both loci as far as phenotype is concerned.

So, in the self cross:
AaBb x AaBb =>
9 A_B_ double dominance = dominance
3 A_bb dominance from A
3 aaB_ dominance from B
1 aabb recessive
So that's where you get your 15:1 dominant:recessive phenotypic ratio.

Now, with a testcross, you were on the right track:
AaBb x aabb =>
1 AaBb
1 Aabb
1 aaBb
1 aabb
So, comparing this to the selfing results above, you'd get 3 dominant : 1 recessive for your phenotypic ratio.

• Let me know if this answers your question! – Asher F. Feb 10 '17 at 16:52