What is difference in scale between a human cell and a virus.
1 answer:
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To calculate the frequency of the heterozygote genotype (Pq) for this gene we must use the Hardy-Weinberg equation ( p2 + 2pq + q2 = 1 ). This equation relies on the Hardy-Weinberg principle, a model in population genetics that states that the frequency of the alleles in a population is never changing, only the combinations (the genotypes) are changing.
If there are only two alleles (variations) of this gene in a population, then their frequencies should add up to 1 (100%). From this, we can calculate the frequency of the q allele.
p +q=1
0,3 +q=1
q= 1-0,3
q= 0,7
Now hat we have the frequency of the q allele we can use the HW equation to calculate the frequency of the heterozygotes.
0,09 + 2pq +0.49= 1
2pq +0,58= 1
2pq= 1-0.58
2pq=0,42
The freqency of the heterozygotes in this population is 0.42
Answer and Explanation:
Punnett squares are illustrations that display the product of hybrid trait crosses by genetic breeding. Established parent genotypes and phenotypes are used and are usually compatible with Mendelian ratios and inheritance. The phnomenon influencing this behaviour is the
- Law of Segregation,
- Law of Independent assortment
- and Dominance
For this cross...
Assuming Widow's peak= dominant
<u> 1. What are the dominant and recessive traits in cross?</u>
Parent generation or P generation:
gametes: Ww → Widow's peak dominant (W), No Widow's peak recessive (w)
<u>2. Is the male homozygous dominant, homozygous recessive, or heterozygous?</u>
Heterozygousity describes the allelic makeup; these traits are made of two different forms of the same gene.
Male: Ww → with the W.peak;
(Thus, with both W and w this is a heterozygous trait)
<u>3. Is the female homozygous dominant, homozygous recessive, or heterozygous?</u>
However, homozygous traits are made up of the same forms of the gene called alleles.
Female: ww → without the W.peak
(Thus, with both w ad w this is a homozygous trait)
<u>1. genotypes and phenotypes for parent and offspring...</u>
<u></u>
Assuming heterozygousity of the male , for the cross
P generation: Ww× ww
Widow's peak ×No Widow's peak
F1 generation:
- genotype: Ww × ww ...(from punnet square)
- phenotype: 2 with the W.peak; 2 without
<em>∴ 50% are genotypically heterozygous while 50% is homozygous , 2 with the W.peak; 2 without </em>
