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Andrew [12]
3 years ago
9

How can two organisms with a dominant trait for a specific feature produce an organism with a recessive trait for the same featu

re? Use the words ""allele"" and ""gamete.""
Biology
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

If the two organisms are heterozygous for the dominant trait.

Explanation:

For every trait, a diploid organism receives two forms of gene called ALLELE, from each parent. Allele is the variant form of a gene.

According to Mendel's law of dominance, for a particular trait, an allele is capable of masking the expression of another allele in a gene. The expressed allele is called the DOMINANT allele while the masked allele is called the RECESSIVE allele. Due to this, an organism can express a dominant trait even in a combined or heterozygous state i.e. different alleles.

When the two heterozygous organisms mate or are crossed, they undergo meiosis and their alleles are separated into GAMETES according to Mendel's law of segregation. For example; an organism with genotype (genetic make-up) Aa will produce gametes with A and a alleles.

Note that, a recessive trait can only be expressed in a homozygous state i.e. same allele. Hence, the two heterozygous organisms will produce gametes containing the recessive allele, which will likely combine to produce a recessive phenotype or trait.

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4. The Rh blood type response system is controlled by the D allele. The genotypes DD and Dd are Rh (Rh positive); dd is Rh- (Rh
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Answer:

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.

EXPLANATION:

The equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1 - This is according to Hardy-Weinberg.

So p+q =1

p2+ 2pq+q2 =1

where p = frequency of dominant allele

q is the frequency of the recessive allele.

from the question, it was given that there were  170 individual out of 400, which were Rh- negative,

So q2 = 170/400 = 0.425

q= 0.65

Also p+q =1

so p = 1-q

or p = 1-0.65

Hence p =0.35

Frequency of homozgupus for D allele = 0.35*0.35 = 0.1225

Frequency of heterozygous or Dd will be 2pq.

or 2*0.35*0.65 = 0.455

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.

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