Question

question 14
two mice contain non - linked genes for two traits. mating two mice that were heterozygous for both traits would produce a phenotype ratio of ______ in their offspring.
a 3:1
b 9:3:4
c 9:3:3:1
d 15:1
e 12:3:1
question 15
1 point
humans have how many autosomes?
a 22 pairs
b 23 pairs
c 44 pairs
d 46 pairs
e 26 pairs
question 16
1 point
which is an example of codominance?

Explanation:

Question 14

Step 1: Determine the genotypes of the parents

Let's assume the two traits are controlled by genes \(A\) and \(B\). The parents are heterozygous for both traits, so their genotypes are \(AaBb\times AaBb\).

Step 2: Use the Punnett - square or the multiplication rule

For a single - gene cross \(Aa\times Aa\), the phenotypic ratio is \(3:1\) (assuming complete dominance). For two non - linked genes \(AaBb\times AaBb\), we use the multiplication rule. The probability of getting the dominant phenotype for the \(A\) - gene is \(P(A -)=\frac{3}{4}\) and for the recessive phenotype \(P(aa)=\frac{1}{4}\). Similarly, for the \(B\) - gene \(P(B -)=\frac{3}{4}\) and \(P(bb)=\frac{1}{4}\).
The four phenotypic classes are:

• \(A - B -\): \(P(A - B -)=\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}\)
• \(A - bb\): \(P(A - bb)=\frac{3}{4}\times\frac{1}{4}=\frac{3}{16}\)
• \(aaB -\): \(P(aaB -)=\frac{1}{4}\times\frac{3}{4}=\frac{3}{16}\)
• \(aabb\): \(P(aabb)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\)

The phenotypic ratio is \(9:3:3:1\)

Question 15

Step 1: Recall the human chromosome number

Humans have a total of \(46\) chromosomes. Chromosomes are divided into autosomes and sex chromosomes.

Step 2: Calculate the number of autosome pairs

There are \(2\) sex chromosomes (\(XX\) in females and \(XY\) in males). The number of autosomes is \(46−2 = 44\). Since chromosomes are in pairs, the number of autosome pairs is \(\frac{44}{2}=22\) pairs

Answer:

Question 14: C. \(9:3:3:1\)
Question 15: A. \(22\) pairs