Please help me on my quiz
1 answer:
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Answer:
The options to this question are unclear but the answer is: Hh, Hh, hh, hh i.e Hh (50%), hh (50%).
Explanation:
This question involves a single Gene coding for the possession or not of Huntington's disease in humans. The disease is said to be an autosomal recessive condition i.e. it only happens in a recessive state (hh).
According to this question, when a female with Huntington disease (hh) mates with a male that is heterozygous (Hh) for the Huntington trait, the following gametes will be produced by each parent.
hh - h and h
Hh - H and h
Using these gametes in a punnet square (see attached image), the following genotypic combination of offsprings will be produced.
Hh, Hh, hh and hh
Hh = 50%
hh = 50%
Huntington Disease is Which percentage
shows the genotype probability *
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>
