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joja [24]
4 years ago
12

In pea plants, purple flower color, C, is dominant to white flower color, c. The table shows the frequencies of the dominant and

recessive alleles in three generations of peas in a garden. Generation 1 has 200 flowers, generation 2 has 400, and generation 3 has 400. Allele Frequency for Flower Color in Peas Generation p q 1 0.5 0.5 2 0.6 0.4 3 0.7 0.3 Which best describes the number of flowers? Generation 1 has more white flowers than generation 2. Generation 3 has more white flowers than generation 2. Generation 2 has fewer purple flowers than generation 1. Generation 1 has fewer purple flowers than generation 3.
Biology
1 answer:
wolverine [178]4 years ago
3 0

Answer:

D

Explanation:

I got it right ;P

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4 0
3 years ago
A couple plans to have a child. the mother is homozygous for an autosomal recessive mutation. the father is heterozygous for the
adoni [48]

<u>Answer</u>:

  • The possible genotypes of the children are Hh and hh.
  • The probability of genotype Hh = 1/2
  • The probability of genotype hh = 1/2

<u>Explanation</u>:

  • <em>Autosomal genes</em> are the genes present on any of the chromosomes except the sex chromosomes.
  • An autosomal recessive mutation is the mutation in an autosomal gene which only expresses itself in the phenotype in homozygous condition and affects both males and female equally.
  • For the given question, let the autosomal gene be represented by H (dominant) and h (recessive)
  • According to the question,

Since the <em>mother</em> is homozygous her genotype would be hh and the <em>father</em> is heterozygous so his genotype would be Hh.

The following cross is made to know the genotypes of the children :

hh X Hh--------> Hh Hh hh hh

So the <em>possible genotypes of the children are Hh and hh as shown in the cross. </em>

Note : An image is attached representing the cross in a Punnett square.

  • The probability of an event = No. of favourable outcomes/total no. of outcomes

The probability of occurrence of genotype Hh would be = 2/4 = 1/2

Similarly, the probability of occurrence of genotype hh would be 2/4 = 1/2



3 0
3 years ago
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