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IB DP Biology: HL复习笔记10.2.2 Skills: Analysing Dihybrid Crosses

Category: IB课程, 教材笔记, 福利干货 Date: 2022年7月27日上午10:21

Predicting Phenotypic and Genotypic Ratios

Worked Example

Fruit flies (Drosophila melanogaster) were crossed in a laboratory study looking for inheritance patterns for two characteristics, wing length and body colour. We can assume that these characteristics are unlinked.

The alleles for these characteristics are as follows:

V = long wings

v = short (vestigial) wings

B = brown body colour

b = black body colour

A black-bodied, heterozygous long-winged fly was crossed with short-winged, homozygous brown-bodied flies. Predict the phenotype ratio of their offspring.

Step 1: Write out the parental genotypes

The first parent is black-bodied and heterozygous long-winged. To be black-bodied it must have the genotype bb. Heterozygous long-winged has the genotype Vv

The second parent is short-winged and homozygous brown-bodied. To be short-winged it must have the genotype vv. Homozygous brown-bodied has the genotype BB

Vvbb × vvBB

Step 2: Identify the gamete genotypes that each parent could produce

These are the allele combinations that each parent can produce in meiosis

Vb vb × vB

Step 3: Complete a Punnett square to show the genotypes of the offspring

Punnett Square showing Genotypes of the Offspring

table-genotypes-of-the-offspring

Step 4: Identify the phenotypes of the offspring

VvBb = long-winged, brown-bodied

vvBb = short-winged, brown bodied

Conclusion: The offspring would be 100% brown-bodied and 1:1 long to short-winged

Test crosses

A test cross can be used to deduce the genotype
The individual in question is crossed with an individual that is expressing the recessive phenotype
This is because an individual with a recessive phenotype has a known genotype
The resulting phenotypes of the offspring provide sufficient information to suggest the genotype of the unknown individual
For a monohybrid test cross:
- If no offspring exhibit the recessive phenotype then the unknown genotype is homozygous dominant
- If at least one of the offspring exhibit the recessive phenotype then the unknown genotype is heterozygous
For a dihybrid test cross:
- If no offspring exhibit the recessive phenotype for either gene then the unknown genotype is homozygous dominant for both genes
- If at least one of the offspring exhibit the recessive phenotype for one gene but not the other, then the unknown genotype is heterozygous for one gene and homozygous dominant for the other
- If at least one of the offspring exhibit the recessive phenotype for both genes then the unknown genotype is heterozygous for both genes

Worked Example

Worked example: Test crosses

Rabbits have a single gene for ear length that has two alleles:
- D, a dominant allele that produces long ears
- d, a recessive allele that produces shorter ears
A breeder has a rabbit called Floppy that has long ears and they want to know the genotype of the rabbit
- There are two possibilities: DD or Dd
The breeder crosses the long-eared rabbit with a short-eared rabbit
- A rabbit displaying the recessive short ear phenotype has to have the genotype dd

Test Cross Possibility Table
Test-Cross-Possibility-One-Table

The predicted ratio of phenotypes of offspring – 1 long ears
The predicted ratio of genotypes of offspring – 1 Dd

Test Cross Possibility Two Table

Predicted ratio of phenotypes of offspring – 1 long ears : 1 short ears
Predicted ratio of genotypes of offspring – 1 Dd : 1 dd
The breeder identifies the different phenotypes present in the offspring
There is at least one offspring with the short ear phenotype
This tells the breeder that their rabbit Floppy has the genotype Dd
If Floppy was genotype DD none of the offspring would have short ears

Worked Example

Worked example: Hard Question

A farmer wishes to maximise his yield of soybean oil from his crop. Oil is extracted from the seeds which typically measure 6-12 mm in diameter. In one species of soybean, Glycine max, two characteristics are governed by the following pairs of unlinked alleles.
H = high oil content in the seeds; h = low oil content in the seeds
E = four seeds in a pod; e = two seeds in a pod
The farmer crossed two soybean plants, both with high oil content and four seeds per pod. This cross resulted in 381 offspring plants being produced. The 381 F₁ offspring had a phenotypic spread as shown in the table below.

table-phenotypic-spread-of-f1-generation-
Use the data in the table to deduce the genotypes of the parent plants in this cross.

Step 1: Write out the possible parental genotypes

Both parents were high-oil content with 4 seeds per pod, which are the dominant traits for each pair of alleles. Neither parent can be homozygous recessive for either oil content or the number of seeds.

Possible allele combinations for oil content: HH, Hh
Possible allele combinations for 4 seeds per pod: EE, Ee
Therefore, the possible parental genotypes were: HHEE, HHEe, HhEE, HhEe

Step 2: Examine phenotype ratios

The smallest number of F₁ offspring was 23 for low oil content, 2 seeds. These must have the homozygous recessive genotype, hhee.The ratios of other phenotypes, relative to the homozygous recessive phenotype, were:
low oil, 4 seeds : hhee = 70:23 = 3.04 : 1 ≅ 3:1
high oil, 2 seeds : hhee = 73:23 = 3.17 : 1 ≅ 3:1
high oil, 4 seeds : hhee = 215:23 = 9.3 : 1 ≅ 9:1

Step 3: Apply Mendel's Law of independent assortment for linked genes

The phenotypes of the offspring display an approximate 9:3:3:1 ratio. This characterises a cross between two double-heterozygous parents where the genes are unlinked, as in this caseConclusion: Both parents' genotype was HhEe / double heterozygous

Exam Tip

Make sure before you start a test cross you think about the following: how many genes are there, how many alleles of each gene are there, which is the dominant allele, what type of dominance is it and is there linkage or codominance between genes?

Even though we learn about fruit flies (Drosophila melanogaster) being studied in detail which led to the discovery of gene linkage, many of their characteristics are unlinked and can be studied on a large scale in this manner.

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