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Art [367]
3 years ago
7

A snapdragon with pink flowers with the genotype Rr is mated with a snapdragon with red flowers with the genotype RR. What is th

e possibility of obtaining a flower with a heterozygous genotype?
The possibility of obtaining a heterozygous flower is %.
Biology
2 answers:
ELEN [110]3 years ago
8 0

The answer would be 50%

yKpoI14uk [10]3 years ago
6 0
<span>50%
 Since one of the parents is homozygous for RR, we absolutely know that parent will be contributing the R allele. And in order to get a heterozygous offspring, it has to have the genotype Rr. In order for the r allele to be contributed it has to come from the genotype Rr parent. And for that parent, half of its offspring will be given the R allele and half the r allele. So there's a 1/2 or 50% probability of getting a heterozygous flower from those 2 parents.</span>
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In Drosophila, the genes crossveinless and Stubble are linked, about 7 map units apart on chromosome 3. cv is a recessive mutant
stealth61 [152]

Answer:

0.035

Explanation:

<u>cv+ is the wild-type dominant allele over cv, therefore:</u>

  • cv+cv+ and cv+cv cause wild-type phenotype for crossveinless
  • cv cv causes the crossveinless phenotype

<u>Sb is a dominant mutant allele over wild-type Sb+, therefore:</u>

  • Sb Sb and Sb Sb+ cause Stubble phenotype
  • Sb+ Sb+ causes wild type phenotype for Stubble

<h3><u>Test cross</u></h3>

It's the cross between the heterozygous female with a homozygous recessive male. Remember that cv and Sb+ are the recessive alleles.

\frac{cv\   Sb^+}{cv^+\ Sb}  X \frac{cv \ Sb+}{cv \ Sb+}

-The male produces only 1 type of gamete: cv Sb+

-The female produces 4 types of gametes:

  • cv Sb+   ] Parental
  • cv+ Sb   ] Parental
  • cv Sb     ] Recombinant
  • cv+ Sb+ ] Recombinant

The genes are linked and separated by 7 map units. A distance of 7 mu means that 7% of the resulting gametes will be recombinant. Because there are 2 possible recombinant gametes, each of them will appear in 3.5% of the cases.

The genotypes and proportions of the offspring resulting from the test cross can be seen in the Punnett Square. The phenotypically wild-type individuals will have the genotype cv+ Sb+ / cv Sb+ (heterozygous for crossveinless and homozygous recessive for Stubble) and a 0.035 proportion.

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