<span>Below are useful in creating a phylogenetic tree of a vertebrate taxon:</span>
<span>1. </span>DNA sequence from living organism or also known as genetic sequence, it is used to determine strand of DNA
<span>2. </span>Behavioral data from living species, a method to identify the origin of individual in behavioral aspect
<span>3. </span><span>Morphological data from fossils, study the arrangement and physical aspect of organism </span>
Answer:
See the answer below
Explanation:
Let the disorder be represented by the allele a.
Since the disease is an autosomal recessive one, affected individuals will have the genotype aa and normal individuals will have the genotype Aa or AA.
Since the four adults are carriers, their genotypes would be Aa.
Aa x Aa
Progeny: AA 2Aa aa
Probability of being affected = 1/4
Probability of being a carrier = 1/2
Probability of not being affected = 3/4
(a) The chance that the child second child of Mary and Frank will have alkaptonuria = 1/2
(b) The chance that the third child of Sara and James will be free of the condition = 3/4
(c)
(d) If someone has no family history of the disorder, their genotype would be AA.
AA x aa
4 Aa
<em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history </em>= 0
(e)
(f) <em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history</em> = 0
Answer:
Crossing over
Explanation:
Crossing over is the process during which two chromatids of two homologous chromosomes exchange part of their genetic segments. It occurs during the pachytene stage of prophase I of meiosis I.
Linked genes are mostly inherited together and do not exhibit independent assortment. However, when linked genes are present far apart from each other on the same chromosome, crossing over can occur between them to produce recombinant chromatids. Therefore, crossing over can break the linkage and produce recombinant progeny as it occurs during the independent assortment of unlinked genes.
the correct answer is B. it could be used to create a complete genomic library.
