Campo J L, Alvarez C
Departamento de Producción Animal, Instituto Nacional de Investigaciones Agrarias, Madrid, Spain.
Poult Sci. 1991 Jan;70(1):1-5. doi: 10.3382/ps.0700001.
The genetic basis for plumage color of the Blue Andalusian breed was studied. Results of crosses between Blue Andalusian females and Brown (eb/eb) tester males showed that this genetic stock was E/E and did not carry a columbian-type gene. This fact was further verified by the cross between Blue Andalusian males and Melanotic Prat (eWh/eWh Co/Co Ml/Ml) females. It is suggested that the Bl/bl+ genotype is effective in changing black to blue pigment when only one eumelanizing gene is present in the genetic background, but it is ineffective in the presence of two different genes producing eumelanin simultaneously. With an E/E genotype, Bl/bl+ does not change black pigment to blue in the areas where the melanotic (Ml) or lacing (Lg) genes produce black pigment, resulting in the laced plumage pattern of the Blue Andalusian (E/E Bl/bl+ Ml-Lg/Ml-Lg). On a non-E/E genetic background, a single dosage of Bl changes the black pigment to blue in the presence of the melanotic or lacing genes. Double-laced phenotypes were not found in either cross, thereby causing the authors to question the role of Co in double and single-laced patterns. Linkage between Ml and Lg was estimated to be 12.2% +/- 2.1 (SE).
对蓝色安达卢西亚鸡品种羽毛颜色的遗传基础进行了研究。蓝色安达卢西亚母鸡与棕色(eb/eb)测试公鸡杂交的结果表明,该遗传群体为E/E,不携带哥伦比亚型基因。蓝色安达卢西亚公鸡与黑化普拉特鸡(eWh/eWh Co/Co Ml/Ml)母鸡的杂交进一步证实了这一事实。研究表明,当遗传背景中仅存在一个真黑素生成基因时,Bl/bl+基因型可有效地将黑色色素转变为蓝色色素,但在同时存在两个产生真黑素的不同基因时则无效。在E/E基因型中,Bl/bl+不会在黑化(Ml)或镶边(Lg)基因产生黑色色素的区域将黑色色素转变为蓝色,从而形成蓝色安达卢西亚鸡(E/E Bl/bl+ Ml-Lg/Ml-Lg)的镶边羽毛图案。在非E/E遗传背景下,单一剂量的Bl会在存在黑化或镶边基因的情况下将黑色色素转变为蓝色。在任何一个杂交组合中均未发现双镶边表型,因此作者对Co在双镶边和单镶边图案中的作用提出了质疑。估计Ml和Lg之间的连锁率为12.2%±2.1(标准误)。