Question

short answer. write the word or phrase that best completes each statement or answers the question on separate sheet of paper.

1. assume that you have a garden and some pea plants have solid leaves and others have striped leaves. you conduct a series of crosses (a through e) and obtain the results given in the table

progeny
cross solid striped
(a) solid x striped 55 60
(b) solid x solid 36 0
(c) striped x striped 0 65
(d) solid x solid 92 30
(e) solid x striped 44 0
define gene symbols and give the possible genotypes of the parents of each cross.

1. the phenotype of vestigial (short) wings (g) in drosophila melanogaster is caused by a mutant gene that independently assorts with a recessive gene for hairy (h) body. the wild - type (gghh x gghh) were crossed among each other to produce 1024 offspring. what phenotypes would you expect among the 1024 offspring, and how many of each phenotype would you expect?
2. a certain type of congenital deafness in humans is caused by a rare autosomal dominant gene.

(a) in a mating involving a deaf man and a deaf woman (both heterozygous), would you expect all the children to be deaf? explain your answer.
(b) in a mating involving a deaf man and a deaf woman (both heterozygous), could all the children have normal hearing? explain your answer.

Explanation:

1.

Step1: Determine dominant - recessive

Let \(S\) represent the allele for solid leaves (dominant) and \(s\) represent the allele for striped leaves (recessive).

Step2: Analyze cross (a)

Solid \(×\) Striped gives a nearly 1:1 ratio. So the parents are \(Ss\) (solid) and \(ss\) (striped).

Step3: Analyze cross (b)

Solid \(×\) Solid gives all solid. Parents could be \(SS×SS\) or \(SS×Ss\).

Step4: Analyze cross (c)

Striped \(×\) Striped gives all striped. Parents are \(ss×ss\).

Step5: Analyze cross (d)

Solid \(×\) Solid gives a 3:1 ratio. Parents are \(Ss×Ss\).

Step6: Analyze cross (e)

Solid \(×\) Striped gives all solid. Parents are \(SS×ss\).

Step1: Determine the cross - genotypes

The cross is \(GgHh×GgHh\). Using the Punnett - square or the product rule, the phenotypic ratio for two independently assorting genes is 9:3:3:1.

Step2: Calculate the number of each phenotype

The total number of offspring is \(n = 1024\).
For the wild - type (\(G\_H\_\)): \(\frac{9}{16}\times1024= 576\).
For vestigial wings and normal body (\(ggH\_\)): \(\frac{3}{16}\times1024 = 192\).
For normal wings and hairy body (\(G\_hh\)): \(\frac{3}{16}\times1024=192\).
For vestigial wings and hairy body (\(gghh\)): \(\frac{1}{16}\times1024 = 64\).

Step1: Determine the cross - genotypes

Let \(D\) be the dominant allele for deafness and \(d\) be the recessive allele for normal hearing. The cross is \(Dd×Dd\).

Step2: Calculate the probability of deaf offspring

Using a Punnett - square, the genotypic ratio is \(DD:Dd:dd=1:2:1\). The probability of a deaf offspring (\(DD\) or \(Dd\)) is \(\frac{3}{4}\), and the probability of a normal - hearing offspring (\(dd\)) is \(\frac{1}{4}\). So we would not expect all children to be deaf.

Step3: Answer part (b)

Since the probability of a normal - hearing offspring (\(dd\)) is \(\frac{1}{4}\), it is possible for all children to have normal hearing, but the probability of this event for multiple children is \((\frac{1}{4})^n\) (where \(n\) is the number of children), which gets smaller as \(n\) increases.

Answer:

(a) \(Ss×ss\)
(b) \(SS×SS\) or \(SS×Ss\)
(c) \(ss×ss\)
(d) \(Ss×Ss\)
(e) \(SS×ss\)

2.