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Anit [1.1K]
2 years ago
8

Using your knowledge of Punnett squares and hybridization, what is the probability that a purebred monohybrid, dihybrid, three-f

actor cross or higher will lead to the expression of dominant phenotypes in first-generation offspring?
Biology
2 answers:
Nimfa-mama [501]2 years ago
8 0

Answer:

100%

Explanation:

No matter how many factors are in the cross, if an 2 purebred (homozygous individuals) are crossed, (one dominant, one recessive) the dominant phenotype will always be displayed.

Imagine a cross with between two individuals true breeding for 6 traits. One shows all dominant genotypes, one shows all recessive genotypes. The only gametes those individuals can pass on will always produce heterozygotes.

AABBCCDDEEFFGG x aabbccddeeffgg

The first individual can only give ABCDEFG alleles. The second individual can only give abcdefg alleles. Therefore, all offspring will be AaBbCcDdEeFfGg, and will therefore express the dominant trait.

Harman [31]2 years ago
8 0

Answer:

1. dihybrid crosses  

2. intermediate  

3. codominance

4.In the first generation of a purebred cross with any number of genes involved, each resulting genotype will be heterozygous for the genes, and the dominant phenotype will, therefore, be expressed in each offspring.

5.In the process of meiosis, alleles separate and are added to the gametes in an independent manner. While each gamete (egg or sperm cell) therefore contains one allele for each gene, it's impossible to know which allele, because alleles segregate in a random manner. However, it's possible to calculate probabilities that the offspring will have a given allele using Punnett squares.

6.Teratogens interrupt the normal developmental process and interfere with pregnancy, resulting in birth defects, even in cases where the parents of the child have normal genetic characteristics. Therefore, teratogens represent a significant environmental factor that could lead to unhealthy and even fatal pregnancy outcomes. To have a healthy pregnancy, women must avoid alcohol, thalidomide, and other substances known to cause birth defects to enable the baby to grow and develop properly.

7.Tigers and cheetahs are other examples of wild animals whose genetic codes include the instructions for elaborate prints on their bodies.

8.Oftentimes, these exotic prints serve as camouflage so the animal can hide from predators. Sadly, such beautiful animals are also poached by hunters who use their furs for clothing and rugs, which has led to many of these wild animals becoming endangered.

Explanation:

PENNFOSTER

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3 years ago
Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advant
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Answer:

0.09

Explanation:

<h3><u>Before selection</u></h3>

Total number in population = 1000

Genotype frequencies

Genotype frequency of BB = 500/1000 = 0.5

Genotype frequency of Bb = 250/1000 = 0.25

Genotype frequency of bb = 250/1000 = 0.25

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.5 + (0.25/2) = 0.625

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.25 + (0.25/2) = 0.375

<h3><u>After selection</u></h3>

We are told that after selection, the genotype frequency of bb is changed as they become 50% less fit. This means the frequency of bb individuals changes from 250 to 125 individuals (50% reduction).

Now the total number of individuals is 500 + 125 + 250 = 875.

Genotype frequencies

Genotype frequency of BB = 500/875 = 0.57

Genotype frequency of Bb = 250/875 = 0.29

Genotype frequency of bb = 125/875 = 0.14

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.57 + (0.29/2) = 0.715

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.14 + (0.29/2) = 0.285

<h3><u>Change in frequency of B after 1 generation</u></h3>

0.715 - 0.625 = 0.09

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