How Do Genetic Traits Determine Physical Characteristics?

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Discussion Overview

The discussion revolves around the genetic determination of physical characteristics, specifically focusing on traits such as eye color and the presence of extra digits. Participants explore methods for determining dominance in genetic traits, calculate probabilities of genetic conditions in offspring, and clarify misconceptions about genetic inheritance.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests crossing a blue-eyed individual with a brown-eyed individual to observe offspring traits to determine dominance.
  • Another participant challenges the claim that Gary's genotype is rr, arguing that having extra digits indicates he must have at least one dominant allele.
  • Participants discuss the use of Punnett squares to calculate probabilities of offspring traits, including the likelihood of having children with phenylketonuria (PKU).
  • One participant presents a Punnett square for the extra digits scenario, estimating a 50% chance of children having extra digits.
  • Another participant calculates the probability of having two unaffected children from a couple with a child affected by PKU, asserting that the events are independent.
  • There is a calculation presented for the probability that both children will not be affected, leading to a numerical estimate of 0.5625.

Areas of Agreement / Disagreement

Participants express differing views on the genetic makeup of individuals with extra digits and the correct interpretation of Punnett squares. There is no consensus on the genotype of Gary, and the calculations regarding probabilities are discussed but not universally agreed upon.

Contextual Notes

Some assumptions about the genotypes of the parents are not explicitly stated, and the discussion includes varying interpretations of genetic dominance and inheritance patterns. The calculations rely on specific allele definitions that may not be universally accepted.

Who May Find This Useful

Individuals interested in genetics, particularly in understanding inheritance patterns and probability calculations related to genetic traits.

colton4286
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1. How could you find out whether blue eyes or brown eyes is dominant? 1. My thoughts: Would you just cross a blue-eyed individual and a brown-eyed indivdual and observe what's prevalent in their offspring?

2. Gary has 6 fingers on each hand & 6 toes on each foot. His wife Sandy and daughter Stephanie are normal. Extra digits is a dominant trait. What fraction of their children would be expected to have extra digits? My thougts: Let's say R= dominant allele, r= recessive allele. Since Gary has extra digits, I know his genotype is rr, but how do I do if the mother is RR or Rr?

3. Phenylketonuria (PKU) is a genetic defect harming the brain. A normal couple have a baby with PKU. What is the probability that their next child will be affected? What is the probability that if they have 2 more children, neither will be affected?
 
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1. You could do that, but I doubt you'd get many volunteers. You could also look at the offspring of a blue-eyed and brown-eyed couple.

2. You're wrong here, rr would mean that Gary has recessive alleles for the gene and would be normal, but having extra digits means that Gary has at least one dominant allele for the trait since having extra digits is a dominant trait.

3. Again, do a Punnett square. See how the offspring differ if one parent is heterozygous and one parent is homozygous dominant, or if both parents are heterozygous. Only one square will allow you to have a child who has PKU where neither parent has it.
 
Thank you for the help. Below is my work. Could someone please double check?

2. Punnett square is Rr X rr. 50% of their children would be expected to have extra digits.
3. Let alleles A=normal, a= has PKU. Punnett square is Aa X Aa.

Probability that their next child will be affected= 25%. Probability that if they have 2 more children neither will be affected= 75%
 
The chance that one child will not be affected is 75%. The chance that another child will not be affected is 75%. (These are independent events).

The chance that BOTH children will not be affected is...
 
0.75 X 0.75 = 0.5625 ?
 

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