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How do you guys approach genetics questions? Here are the ones I know:
1) looking for distinct gametes when given a genotype:
Use the 2^n rule, where n equals the number of heterozygous alleles.
2)looking for distinct offspring when given the genotype of each parent:
Find the number of different types of possible alleles for each gene and multiply them together. Ex. If parrents have AaBb and AaBb, you have 3 different for A and 3 different for B, so just multiply them together and you get 9 different possible distinct offspring.
3)If autosomal recessive genes are 1/50 in the population, and both parents have it, then what is the probability that the child is homozygous recessive?
For this you multiply 1/50 *1/50 * 1/4 (probability of child being homozygous recessive)
4) If colorblindness is present in 1/12 of men. How frequently does it occur in women? The solution is 1/12*1/12, as woman needs both genes - from mom and dad - to be colorblind (as colorblindness is x-linked recessive)
5) (got this from another thread) If you have two parents who are heterozygous carriers for an AR mutation and they give birth to a healthy child, what is the probability that child is homozygous normal?
Aa x Aa = AA, Aa, Aa, aa. Now since the kid is normal, it (yes, it!) can't be aa, so it is 1/3.
Ok, I kinda still suck at genetics problems, prob cause I never did any outside of bio 1. Post any other problems with solutions!
1) looking for distinct gametes when given a genotype:
Use the 2^n rule, where n equals the number of heterozygous alleles.
2)looking for distinct offspring when given the genotype of each parent:
Find the number of different types of possible alleles for each gene and multiply them together. Ex. If parrents have AaBb and AaBb, you have 3 different for A and 3 different for B, so just multiply them together and you get 9 different possible distinct offspring.
3)If autosomal recessive genes are 1/50 in the population, and both parents have it, then what is the probability that the child is homozygous recessive?
For this you multiply 1/50 *1/50 * 1/4 (probability of child being homozygous recessive)
4) If colorblindness is present in 1/12 of men. How frequently does it occur in women? The solution is 1/12*1/12, as woman needs both genes - from mom and dad - to be colorblind (as colorblindness is x-linked recessive)
5) (got this from another thread) If you have two parents who are heterozygous carriers for an AR mutation and they give birth to a healthy child, what is the probability that child is homozygous normal?
Aa x Aa = AA, Aa, Aa, aa. Now since the kid is normal, it (yes, it!) can't be aa, so it is 1/3.
Ok, I kinda still suck at genetics problems, prob cause I never did any outside of bio 1. Post any other problems with solutions!
