Question

Hello, please please please answer these questions! I would appreciate it!!!

Questions 2 to 4 are based on the following scenario Question 2: Although the following scenario is hypothetical, it is based on actual facts. Pukaha National Wildlife Centre is a captive breeding facility located in a protected forest area in the lower part of the North Island of Aotearoa, New Zealand. It aims to restore native wildlife, with its initial target on birds. One of its breeding programs focuses on the North Island Brown Kiwi whose current conservation status is classed as vulnerable. In 2011, the first white kiwi chick hatched at the wildlife centre. This was very unexpected as North Island Brown Kiwis, as the name implies, have brown feathers. This white kiwi did not have albinism since it did produce pigmentation, it was simply a North Island Brown Kiwi with white feathers. White kiwis have been observed in some wild populations, although this phenotype is very rare. This gene for kiwi feather colour follows Mendelian genetics, whereby, the allele for brown feathers is dominant and the allele for white feathers is recessive. In order for a kiwi to have white feathers both its parents must carry this rare white allele and both pass that on to that offspring. Since this initial white kiwi hatched at this facility, two more have been born. ***** You have decided to travel to New Zealand and volunteer in the kiwi breeding program. You had just finished BIOL2050H before going overseas, so you were eager to apply your genetic analytical skills on this population. The following table summaries the phenotype distribution for this population of North Island Brown Kiwi currently at the wildlife centre:

Thank you soooo much!!!!

Answer 1

According to the rules of the company, I can answer only first question. I am answering it in detail so that you understand it properly.

Genotypic frequency of a genotype = number of individuals with that genotypes / total number of individuals

Allele frequency = √genotypic frequency = genotypic frequency of that genotype + 1/2 genotypic frequency of heterozygotes

= 10.9118 Brown phenotype = BB + B6 het = 10.9/16 white prenotype = 60 a = (0.0683) 10.0883 (gano la nec frequnay sa you oor 90.0883 (96JO0883 = 10:298 . Biz 1-6 = 1-0-298 . (p) = 10-102]

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