We read your photo as:

Carnation plants show co-dominance for the anthocyanin gene. FA – allele for anthocyanin pigment (red flowers) FN – allele for no anthocyanin pigment (white flowers). State the genotype of a carnation

Biology

ScanSolve AI Answer

Answers

3 parts
c(i)

FAFNF^A F^N

c(ii)

Red and white flowers\text{Red and white flowers}

c(iii)

24 or 12\frac{2}{4} \text{ or } \frac{1}{2}

Here's the breakdown of the genetics problems:

c) Carnation plants show co-dominance for the anthocyanin gene.
FAF^A – allele for anthocyanin pigment (red flowers)
FNF^N – allele for no anthocyanin pigment (white flowers)

i) State the genotype of a carnation plant that is heterozygous for this gene.
A heterozygous plant has two different alleles.
The genotype is FAFN\boxed{F^A F^N}.

ii) Describe the phenotype of a heterozygous carnation plant for this gene.
Since the alleles show co-dominance, both pigments are expressed.
The phenotype is Red and white flowers\boxed{\text{Red and white flowers}}.

iii) The breeder crossed a FAFNF^A F^N carnation plant with a FAFAF^A F^A carnation plant. Predict, using a genetic diagram, the proportion of pure breeding carnation plants in the offspring.

Parental genotypes: FAFN×FAFAF^A F^N \times F^A F^A

Gametes:
For FAFNF^A F^N: FAF^A, FNF^N
For FAFAF^A F^A: FAF^A, FAF^A

Punnett square:

FAFNFAFAFAFAFNFAFAFAFAFN\begin{array}{|c|c|c|} \hline & F^A & F^N \\ \hline F^A & F^A F^A & F^A F^N \\ \hline F^A & F^A F^A & F^A F^N \\ \hline \end{array}

Offspring genotypes:
FAFAF^A F^A, FAFNF^A F^N

Offspring phenotypes:
Red flowers (FAFAF^A F^A), Red and white flowers (FAFNF^A F^N)

Proportion of pure breeding carnation plants:
Pure breeding plants are homozygous. In this cross, the homozygous genotype is FAFAF^A F^A.
There are 2 FAFAF^A F^A offspring out of 4 total offspring.
The proportion is 24 or 12\boxed{\frac{2}{4} \text{ or } \frac{1}{2}}.

3) a) Sex in cats is determined in the same way as in humans. Complete the diagram below to show how sex is determined in cats.

Parental genotypes: Male cat (XYXY) ×\times Female cat (XXXX)

Gametes:
Male cat: XX, YY
Female cat: XX, XX

Punnett square:

XYXXXXYXXXXY\begin{array}{|c|c|c|} \hline & X & Y \\ \hline X & XX & XY \\ \hline X & XX & XY \\ \hline \end{array}

Offspring genotypes:
XXXX, XYXY

Offspring phenotypes:
Female, Male

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question