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sladkih [1.3K]
3 years ago
9

Red-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son.

What are the genotypes of the parents?
Biology
1 answer:
Alborosie3 years ago
6 0

Genotype of Mother: XX°

Genotype of Father: XY

Explanation:

  • Since the Red-green colour blindness is a recessive trait , it will not be visible in a heterozygous female because dominant allele will mask the recessive allele.
  • Father has only one X chromosome, thus if the trait was present in  that X chromosome he must have shown the trait because all the alleles in Y chromosome are recessive.
  • Since the son has inherited the trait it is obvious that he had recieved the trait from his mother who must have been the carrier.
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A population of 100 tigers has 30 with horizontal stripes. Using pedigrees, you ascertain that horizontal striping is caused by
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Assuming that this population is in Hardy-Weinberg equilibrium, we can sayt that  <em>the</em><em> probability</em><em> that, given enough time, all tigers in the population will have horizontal stripes is</em><em> 30% = 0.3.</em>

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<u>Available data:</u>

  • Population size, N = 100 tigers
  • Number of Tigers with horizontal stripes = 30
  • Horizontal stripes is caused by an autosomal recessive mutation.

We need to know the probability that all tigers in the population will have horizontal stripes, which is the recessive phenotype.

We assume that the population is in Hardy-Weinberg equilibrium, so there should be no evolution.

Since the population is in H-W equilibrium, the allelic, genotypic and phenotypic frequencies will remain the same generation after generation.

Now, we will calculate the allelic and genotypic frequencies of horizontal striped individuals in the population.

There are 100 individuals, and only 30 have horizontal stripes.

So, the phenotypic frequency, F(HS) is 30/100 = 0.3 = 30%

Since this is a recessive phenotype, this value equals the genotypic frequency, F(hh) = 30%

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F(hh) = q² = 0.3

f(h) = q = √0.3 = 0.55 = 55%

According to these calcs, the probability for the fixation of the recessive allele is 55%, and the probability that all individuals express the recessive horizontal stripes phenotype is 30%.

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You can learn more about Hardy-Weinberg equilibrium at

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