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In Drosophila, an autosomal gene determines the shape of the hair, with \(B\) giving straight and \(b\) giving bent hairs. On another autosome, there is a gene of which a dominant allele \(I\) inhibits hair formation so that the fly is hairless ( \(i\) has no known phenotypic effect). a. If a straight-haired fly from a pure line is crossed with a fly from a pure-breeding hairless line known to be an inhibited bent genotype, what will the genotypes and phenotypes of the \(\mathrm{F}_{1}\) and the \(\mathrm{F}_{2}\) be? b. What cross would give the ratio 4 hairless: 3 straight: 1 bent?

Short Answer

Expert verified
F1: all hairless (genotype \( BbIi \)); F2: 9 hairless, 6 straight, 1 bent. Cross \( BbIi \) with \( Bbii \) for a 4:3:1 ratio.

Step by step solution

01

Understanding Parental Genotypes

First, let's identify the genotypes of the parents. The straight-haired fly from a pure line is homozygous for the straight hair gene, so it has the genotype \( BB \). Because it has hair, it must also be \( ii \) for the inhibitor gene. The hairless fly is from a line that has bent hair inhibited by dominance. Therefore, it has the genotype \( bbII \).
02

Determining the F1 Generation

Cross the genotypes \( BBii \) (straight-haired) with \( bbII \) (hairless). The resulting \( F_1 \) generation will have the genotype \( BbIi \). Since \( I \) is dominant, all \( F_1 \) flies will be hairless.
03

Setting up the F2 Generation Cross

To find the \( F_2 \) generation, cross two \( F_1 \) individuals, both of which have the genotype \( BbIi \).
04

Calculate F2 Phenotypic and Genotypic Ratios

Using a Punnett square, calculate the genotypes. The genotypic ratios are as follows: 1/16 \( BBIi \), 2/16 \( BbIi \), 1/16 \( bbIi \) (all hairless due to \( I \)), 2/16 \( BBii \) (straight), 4/16 \( Bbii \) (straight), and 1/16 \( bbii \) (bent). The phenotypic ratio is 9 hairless (due to any \( I \)) : 6 straight (\( ii \) and any \( B \) allele) : 1 bent (\( bbii \)).
05

Determine Cross for Desired Ratio

To achieve a 4:3:1 ratio of hairless to straight to bent, you need a cross where \( I \), dominant over any hair form, is not fully expressed. The cross could be \( BbIi \) with \( Bbii \), resulting in an approximately 4:3:1 phenotypic ratio by dominant interactions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Autosomal Genes
Autosomal genes are genes located on any of the numbered, non-sex chromosomes in an organism. In fruit flies (Drosophila), autosomal genes play a crucial role in determining physical characteristics that are inherited in a straightforward manner. These genes are present in pairs, one inherited from each parent. They determine traits such as wing size, eye color, and, as in our exercise, hair shape.

In this scenario, the gene that influences hair shape (straight or bent) is autosomal. This means the gene in question resides on one of the many non-sex chromosomes in the Drosophila genome. Thus, it's inherited regardless of the fly's sex, following typical Mendelian inheritance patterns.

Autosomal genes represent a "normal" pattern of inheritance, contrasting with genes found on sex chromosomes (X or Y), which can behave differently due to the unequal number of these chromosomes in males and females.
Dominant Allele
A dominant allele is a version of a gene that "dominates" over others within the same gene pair to determine a trait. When you have two different alleles, the dominant one masks the effect of the recessive allele and will be the characteristic that appears in the organism.

In the context of our Drosophila exercise, the allele for hair inhibition, represented by the letter "I," is dominant. This means that even if a fly inherits only one of these "I" alleles, it will prevent hair formation, leading the fly to be hairless regardless of its other hair-shape allele ("B" for straight or "b" for bent).

It's important to remember that a dominant allele doesn't "do more"; it simply masks the presence of a recessive allele in terms of visible traits. In this case, it doesn't matter whether the fly has straight or bent hair alleles; as long as the dominant "I" is present, the fly will be hairless.
Punnett Square
A Punnett square is a useful tool in genetics that helps to predict the genotype and phenotype ratios of offspring from two parents. It lays out all possible combinations of parental alleles, revealing how genetic traits might be inherited.

For our Drosophila example, setting up a Punnett square allows us to visualize the genetic combinations resulting from crossing different genotypes, like a "BbIi" fly from the \( F_1 \) generation with another "BbIi" fly.

This method systematically showcases the inheritance patterns, displaying the probability of offspring inheriting specific trait combinations. For instance, it helps us confirm why out of 16 possible offspring in the \( F_2 \) generation, those who receive at least one "I" allele are hairless due to the dominance of "I."
  • 9/16 flies are hairless (have "I").
  • 6/16 are straight-haired (must be "ii" and have "B").
  • 1/16 has bent hair ("bbii").
This structured prediction aids significantly in understanding genetic inheritance and phenotypic expression.
Phenotypic Ratio
In genetics, the phenotypic ratio is the relative frequency of different phenotypes (or physical expressions of genetic traits) in a population of organisms. It's calculated based on the genotypic ratio, which involves counting the number of each genotype from the offspring of a particular cross.

From our Drosophila inheritance exercise, we derived a phenotypic ratio in the \( F_2 \) generation. The ratio was 9 hairless: 6 straight: 1 bent from the cross of two "BbIi" flies. This means:
  • 9 offspring have the "I" allele, making them hairless.
  • 6 offspring have straight hair as they carry "ii" and at least one "B."
  • 1 offspring has bent hair with the genotype "bbii."
Phenotypic ratios provide a straightforward way to look at the visual outcomes (the phenotypes) genetic crosses produce. They allow scientists and students to anticipate how frequently a given trait will appear among offspring.

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Most popular questions from this chapter

A dominant allele H reduces the number of body bristles that Drosophila flies have, giving rise to a "hairless" phenotype. In the homozygous condition, \(H\) is lethal. An independently assorting dominant allele Shas no effect on bristle number except in the presence of \(H,\) in which case a single dose of \(S\) suppresses the hairless phenotype, thus restoring the hairy phenotype. However, \(S\) also is lethal in the homozygous (S/S) condition. a. What ratio of hairy to hairless flies would you find in the live progeny of a cross between two hairy flies both carrying \(H\) in the suppressed condition? b. When the hairless progeny are backcrossed with a parental hairy fly, what phenotypic ratio would you expect to find among their live progeny?

In a maternity ward, four babies become accidentally mixed up. The \(A B O\) types of the four babies are known to be \(\mathrm{O}, \mathrm{A}, \mathrm{B},\) and \(\mathrm{AB}\). The \(\mathrm{ABO}\) types of the four sets of parents are determined. Indicate which baby belongs to each set of parents: (a) \(A B \times O,(b) A \times O,(c) A \times A B\) (d) \(\mathrm{O} \times \mathrm{O}\)

Several mutants are isolated, all of which require compound G for growth. The compounds (A to E) in the biosynthetic pathway to G are known, but their order in the pathway is not known. Each compound is tested for its ability to support the growth of each mutant (1 to 5). In the following table, a plus sign indicates growth and a minus sign indicates no growth. $$\begin{array}{rcccccc} {5}{c} {\text {compound tested}} \\ \ { 2 - 6 }{1}{c} {} & &\mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{E} & \mathrm{G} \\ \hline \text { Mutant } & 1 & \- & \- & \- & \+ & \- & \+ \\ & 2 & \- & \+ & \- & \+ & \- & \+ \\ & 3 & \- & \- & \- & \- & \- & \+ \\ & 4 & \- & \+ & \+ & \+ & \- & \+ \\ & 5 & \+ & \+ & \+ & \+ & \- & \+ \\ \hline \end{array}$$ a. What is the order of compounds A to Ein the pathway? b. At which point in the pathway is each mutant blocked? c. Would a heterokaryon composed of double mutants 1,3 and 2,4 grow on a minimal medium? Would 1,3 and \(3,4 ?\) Would 1,2 and 2,4 and \(1,4 ?\)

On a fox ranch in Wisconsin, a mutation arose that gave a "platinum" coat color. The platinum color proved very popular with buyers of fox coats, but the breeders could not develop a pure-breeding platinum strain. Every time two platinums were crossed, some normal foxes appeared in the progeny. For example, the repeated matings of the same pair of platinums produced 82 platinum and 38 normal progeny. All other such matings gave similar progeny ratios. State a concise genetic hypothesis that accounts for these results.

If a man of blood-group AB marries a woman of bloodgroup A whose father was of blood-group \(\mathrm{O}\), to what different blood groups can this man and woman expect their children to belong?

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