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Levart [38]
3 years ago
5

Brown (B) hair is dominant over Blonde (b) hair. If a person who is BB has children with a person who is BB, what would be their

children’s possible genotypes?
a
BB, Bb
b
bb, BB
c
bb
d
BB
Biology
1 answer:
Shalnov [3]3 years ago
6 0

Answer:

d. BB

Explanation: The only possible phenotype would be BB because if you draw a Punnett square you see that you get BB, BB, BB, and BB so that is the only posssible genotype.

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ValentinkaMS [17]

Answer:

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Cystic fibrosis is most common in individuals of Northern European descent, affecting 1 in 3200 newborns. Assuming that these al
IgorLugansk [536]

Answer:

0.0177

Explanation:

Cystic fibrosis is an autosomal recessive disease, thereby an individual must have both copies of the CFTR mutant alleles to have this disease. The Hardy-Weinberg equilibrium states that p² + 2pq + q² = 1, where p² represents the frequency of the homo-zygous dominant genotype (normal phenotype), q² represents the frequency of the homo-zygous recessive genotype (cystic fibrosis phenotype), and 2pq represents the frequency of the heterozygous genotype (individuals that carry one copy of the CFTR mutant allele). Moreover, under Hardy-Weinberg equilibrium, the sum of the dominant 'p' allele frequency and the recessive 'q' allele frequency is equal to 1. In this case, we can observe that the frequency of the homo-zygous recessive condition for cystic fibrosis (q²) is 1/3200. In consequence, the frequency of the recessive allele for cystic fibrosis can be calculated as follows:

1/3200 = q² (have two CFTR mutant alleles) >>  

q = √ (1/3200) = 1/56.57 >>

- Frequency of the CFTR allele q = 1/56.57 = 0.0177  

- Frequency of the dominant 'normal' allele p = 1 - q = 1 - 0.0177 = 0.9823

4 0
3 years ago
Biochemical engineers have designed a lamp that uses algae to create light. these algae are able to glow in the dark. how could
Tomtit [17]
I'm not sure I would say B

3 0
3 years ago
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Answer:

Explanation:

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c i did this test and got it right

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