After back-crossing, the resulting animal (F1) is heterozygous for the m allele.
ID: 151448 • Letter: A
Question
After back-crossing, the resulting animal (F1) is heterozygous for the m allele. If the mutation m is dominant, both homozygous (m/m) and heterozygous (m/+) F2 animals will produce the M phenotype. How can you make a homozygous mutant animal in this context? Select all that apply.
Cross the F2 M phenotype hermaphrodites with F2 M phenotype males. All M phenotype individuals in the next generation will be m/m
Cross the F2 M phenotype hermaphrodites with F1 males. All M phenotype individuals in the next generation will be m/m
Self the F2 M phenotype hermaphrodites. m/m individuals will produce ¼ m/m offspring with the M phenotype
Self the F2 M phenotype hermaphrodites. m/m individuals will produce only m/m offspring with the M phenotype
Self the F2 M phenotype hermaphrodites. m/+ individuals will produce ¼ wild-type offpring.
Additional back-crosses to wild-type will increase the likelihood of obtaining m/m individuals with the M phenotype from those crosses.
Cross the F2 M phenotype hermaphrodites with F2 M phenotype males. All M phenotype individuals in the next generation will be m/m
Cross the F2 M phenotype hermaphrodites with F1 males. All M phenotype individuals in the next generation will be m/m
Self the F2 M phenotype hermaphrodites. m/m individuals will produce ¼ m/m offspring with the M phenotype
Self the F2 M phenotype hermaphrodites. m/m individuals will produce only m/m offspring with the M phenotype
Self the F2 M phenotype hermaphrodites. m/+ individuals will produce ¼ wild-type offpring.
Additional back-crosses to wild-type will increase the likelihood of obtaining m/m individuals with the M phenotype from those crosses.
Explanation / Answer
Additional back-crosses to wild-type will increase the likelihood of obtaining m/m individuals with the M phenotype from those crosses.
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